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\centerline {\bf ON SOME IDENTITIES FOR THE ELEMENTS}
\centerline {\bf OF A SYMMETRIC MATRIX}
\vskip 12pt
\centerline {\sc N. V. Ilyushechkin}
\vskip 40pt
\centerline {\bf Introduction}
\vskip 12pt

The following problem has its origin in the theory of continuous media
(see [1]).
\medskip

Let ${\rm Sym}(n)$ be the space of $n$-dimensional real symmetric
matrices. Describe all indefinitely smooth maps
$$ {\cal F}: {\rm Sym}(n) \longrightarrow {\rm Sym}(n), $$
possessing the following property: for any matrix $X \in {\rm Sym}(n)$ and
any orthogonal matrix $C$, such that $C^{-1} X C$ is diagonal,
the matrix $C^{-1}{\cal F}(X)C$ is also diagonal.
\medskip

Another problem, in a sense dual to the first one,
consists in describing the set of all infinitely smooth maps
$$ {\cal G}: {\rm Sym}(n) \longrightarrow {\rm Sym}(n), $$
with the property that for any matrix $X \in {\rm Sym}(n)$ and
any orthogonal matrix $C$, such that $C^{-1} X C$ is diagonal,
all the diagonal elements of the matrix $C^{-1}{\cal G}(X)C$
are equal to zero.
\medskip

In this paper, we will prove the following theorems.
\medskip

{\sc Theorem 1.} {\it If} $ {\cal F} $ {\it is a map from the first set,
then there are infinitely smooth functions}
$$ a_m: {\rm Sym}(n) \longrightarrow R $$
{\it for} $ m = 0,1,\ldots,n - 1 $, {\it such that}
$$ {\cal F}(X) = a_0 (X) E + a_1 (X) X + \cdots + a_{n-1} (X) X^{n-1},
   \eqno (0.1) $$
{\it that is} $ {\cal F}(X) $ {\it is a matrix polynomial of degree}
$ n - 1 $ {\it with variable scalar coefficients.}
\medskip

{\sc Theorem 2.} {\it For any map} $ {\cal G} $ {\it from the second set
there is an infinitely smooth skew-symmetric matrix}
$ A(X) $ {\it such that}
$$ {\cal G}(X) = A(X) X - X A(X), \eqno (0.2) $$
{\it that is} $ {\cal G}(X) $ {\it is the commutator of} $ X $
{\it with a variable skew-symmetric matrix.}
\medskip

The proof of these theorems relies on some matrix identities, notably,
a formula for the volume of a parallelepiped in the space of normal
matrices and certain formulas related to the discriminants of the
characteristic polynomials.
These formulas represent independent interest, and we start the paper
by establishing them.

In section 1, we will discuss the volumes generated by functions
of a normal matrix. Section 2 is about discriminant polynomials.
Section 3 contains a proof of one of the identities stated in Section 2
as well as some useful formulas arising from the Hodge isomorphism.
Finally, in Section 4 we give the proof of the two main theorems.

\vskip 24pt
\centerline {\bf \S 1. On volumes generated by functions of
a normal matrix}
\vskip 12pt

Although this study is primarily devoted to symmetric matrices, the formula
we are going to prove in this section is valid in the more general case of
normal matrices with complex elements.

Let ${\rm Mat}_C(n)$ be the space of square complex
matrices of order $n$. It is a unitary space of
dimension $n^2$ with the scalar product of two matrices 
$A=(a_{ij})$ and $B=(b_{ij})$ given by
$$
   (A,B) = \sum_{i,j=1}^n a_{ij}\bar{b}_{ij}.
$$

 Let $Z \in {\rm Mat}_C(n)$ be a normal
matrix, and let $f_1(\zeta),\dots,f_n(\zeta)$ be complex-valued analytic
functions of a complex argument defined on the spectrum ${\lambda}_1,
\dots,{\lambda}_n$ of the matrix $Z$. We are interested in the volume
of the parallelepiped in ${\rm Mat}_C(n)$ built on the vectors
$f_1(Z),\dots,f_n(Z)$.

We begin with the following observation. Let
${\bf w}_1,\dots,{\bf w}_l$ be vectors of a unitary space of dimension
$m \ge l$. Denote by $V({\bf w}_1,\dots,{\bf w}_l)$ the volume of the
parallelepiped built on these vectors.
Viewing all ${\bf w}_j$ as vector-columns
${\bf w}_j = (w_{1j},\dots,w_{mj})^T$, we
can arrange their coordinates into a matrix $ W = (w_{ij}) $, where $i = 1,
\dots,m $ and $ j = 1,\dots,l $. Denote by
$ \mathfrak{M}_r(W) $ the sum of squares of absolute values of all
order $r$ minors of the matrix $W$. It is known [2] that
$$ V^2({\bf w}_1,\dots,{\bf w}_l) = \mathfrak{M}_l(W) \eqno (1.1) $$

Observe that the volume does not change  under the action
of a unitary transformation $Q$ on all the vectors $ {\bf w}_j $:
$$ V({\bf w}_1,\dots,{\bf w}_l) =
              V(Q{\bf w}_1,\dots,Q{\bf w}_l) \eqno (1.2) $$
Indeed, the square of the volume $V^2({\bf w}_1,\dots,{\bf w}_l)$ coincides
with the Gramian generated by the vectors $ {\bf w}_1,\dots,{\bf w}_l $,
and scalar products $ ({\bf w}_j,{\bf w}_k) = (Q{\bf w}_j,Q{\bf w}_k) $ which
are the elements of the Gram matrix, are preserved under unitary
transformations of the whole space.

We return to the evaluation of the volume $ V(f_1(Z),\dots,f_n(Z)) $. From
the elements of matrices $ f_1(Z),\dots,f_n(Z) $ we make up
a new matrix $ \Phi(Z) $ of size $ n^2 \times n $, in which column
number $\nu = 1,\dots,n$ consists of the elements of the matrix
$f_\nu(Z)$ taken in a certain order,
same for all $\nu$. Due to (1.1), we have
$$ V^2(f_1(Z),\dots,f_n(Z)) = \mathfrak{M}_n(\Phi(Z)) \eqno (1.3) $$

The similarity transformation $ N = U^{-1}MU $, where
$M,N$ are variable matrices, while $U$ is
a constant unitary matrix, can be viewed as a unitary transformation
of the space $ {\rm Mat}_C(n) $. We can choose the unitary matrix $ U $
so that the matrix $ L = U^{-1} Z U $ be diagonal:
    $$ L = {\rm diag}(\lambda_1,\dots,\lambda_n) $$
Then, of course, $ f_{\nu}(L) = U^{-1}f_{\nu}(Z)U $ and
$$ f_{\nu}(L) = {\rm diag}(f_{\nu}(\lambda_1),\dots,
f_{\nu}(\lambda_n)) $$

Since the similarity transformation is unitary, (1.2) implies that
$$ V(f_1(Z),\dots,f_n(Z)) = V(f_1(L),\dots,f_n(L)), $$
and owing to (1.3), to determine the unknown volume we only
need to find $\mathfrak{M}_n(\Phi(L))$. The matrix $ \Phi(L) $
contains only one nonzero minor of order $n$ made of diagonal
elements of matrices $ f_{\nu}(Z) $, and $\mathfrak{M}_n(\Phi(L))$
equals the square of the absolute value of this minor. Extracting
the square root, we receive the final formula.
\medskip

{\bf Proposition 1.}
{\it If $Z$ is a complex normal matrix and $f_1$, \dots, $f_n$
are analytic functions defined on the spectrum of $Z$, then
the volume of the parallelepiped spanned by the vectors
$f_1(Z)$, \dots, $f_n(Z)$ in the space of matrices is given by the
formula};
$$ V(f_1(Z),\dots,f_n(Z)) = \vert \left \vert \matrix {
f_1(\lambda_1) & f_2(\lambda_1) & \dots & f_n(\lambda_1) \cr
f_1(\lambda_2) & f_2(\lambda_2) & \dots & f_n(\lambda_2) \cr
\vdots & \vdots & \ddots & \vdots \cr
f_1(\lambda_n) & f_2(\lambda_n) & \dots & f_n(\lambda_n)
} \right \vert \vert $$
\medskip

For example, suppose that all functions $f_{\nu}(\zeta)$ are polynomials
$p_{\nu}(\zeta)$ of degree $n-1$:
$$ p_\nu(\zeta) = b_{\nu 0} + b_{\nu 1}\zeta +
    \dots + b_{\nu, n - 1}\zeta^{n - 1}.
$$
The coefficients of these polynomials make up a square matrix of order $n$:
$$ B = \pmatrix {
b_{10} & b_{11} & \dots & b_{1,n - 1} \cr
b_{20} & b_{21} & \dots & b_{2,n - 1} \cr
\vdots & \vdots & \ddots & \vdots \cr
b_{n0} & b_{n1} & \dots & b_{n,n - 1}} $$
Then, by the determinant multiplication rule,
$$ V(p_1(Z),\dots,p_n(Z)) = \vert {\rm det}(B) \vert \vert
W(\lambda_1,\dots,\lambda_n) \vert, $$
where $W$ is the Vandermonde determinant corresponding to numbers
$ \lambda_1,\dots,\lambda_n $. Raising this in the square, we get
$$ V^2(p_1(Z),\dots,p_n(Z)) = \vert {\rm det}(B) \vert^2
\vert {\rm Dis}(\chi_Z) \vert, $$
where $\chi_Z$ is the characteristic polynomial of
matrix $Z$, and ${\rm Dis}(\chi_Z)$  is its
discriminant. In particular,
$$ V^2(E,Z,\dots,Z^{n - 1}) = \vert {\rm Dis}
(\chi_Z) \vert $$

Taking into account (1.3), we obtain the following algebraic analog
of the previous formula.
\medskip

{\bf Proposition 2.}
{\it If $\chi_Z$ is the characteristic polynomial of a complex normal 
matrix $Z$, and ${\rm Dis}$ its discriminant, then}
$$ \vert {\rm Dis}(\chi_Z) \vert =
     \mathfrak{M}_n (Z_*), \eqno (1.4) $$
{\it where $Z_*$ is the matrix of size $ n^2\times n $, whose column
number $\nu$ contains the elements of the matrix $Z^{\nu}$,
arranged in a definite order, same for all $\nu = 0,1,\dots,n - 1$.}
\medskip

In the case when the normal matrix $ Z = X \in {\rm Sym}(n) $ is symmetric,
formulas (1.3) and (1.4) can be slightly modified.
This modification is related to the fact that the dimension of the space
$ {\rm Sym}(n) $, which we shall denote by $ N = C^2_{n+1} $, is less than
the dimension $n^2$ of the space ${\rm Mat}_C(n)$.

Passing to exact definitions, let us
view a matrix $ X = (x_{ij}) \in {\rm Sym}(n) $
as a column $ \widehat X $ of dimension
$ N $, whose elements are $ x_{ii} $ and $ \sqrt 2 x_{ij} $ with
$ i < j $, arranged in the order
$$ 11,22,\dots,nn,12,13,\dots,1n,23,\dots,2n,\dots,n-1 \: n \eqno (1.5) $$
The choice of the factor $\sqrt2 $ is stipulated by the fact that
the similarity transformation
$$ X = C^{-1} Y C, \eqno (1.6) $$
where $ X $ and $ Y $ are variable symmetric matrices, and $ C $
is a constant orthogonal matrix, preserves the quadratic form
$$ \sum\limits_{i=1}^n x^2_{ii} + 2 \sum\limits_{i<j} x^2_{ij}. $$

By $E_{ij}$ we denote the square matrix of order $ n $,
which has number 1 on the intersection of $i$'th row and $j$'th column
and has zeros elsewhere.
Define the symmetric matrices $ \Gamma_{ii} $ with $ i = 1,\dots,n $ and
$ \Gamma_{ij} $ with $ i < j $ by the equalities
$$ 
  \Gamma_{ii} = E_{ii}, \qquad \Gamma_{ij} = \frac{1}{\sqrt 2}
  (E_{ij} + E_{ji}). \eqno (1.7)
$$
Then $ {\rm Sym}(n) $ can be considered as a Euclidean space of
dimension $ N $ with orthonormal basis (1.7), arranged in the order (1.5).

Slightly modifying the argument used above, we can deduce the formulas
similar to (1.3) and (1.4), in dimension $N$. Instead of the
matrix $ \Phi (X) $ from (1.3), we use the $N\times n$  matrix
$ \Phi_s (X) $ whose $\nu$'th column, $\nu = 1,\dots,n$, contains the
elements of the matrix $f_{\nu}(X)$, arranged in the order (1.5) and
multiplied, where necessary, by $\sqrt2 $. Then formula (1.3) transforms to
  $$ V^2(f_1(X),\dots,f_n(X)) = \mathfrak{M}_n(\Phi_s (X)), $$
and formula (1.4) turns into the assertion of the following proposition.
\medskip

{\bf Proposition 3.}
{\it The discriminant of the characteristic polynomial of a complex
symmetric matrix $X$ is equal to the sum of squares of absolute values
of all $n\times n$ minors of an auxiliary matrix $X_!$:
  $$ {\rm Dis}(\chi_X) =
     \mathfrak{M}_n (X_!), \eqno (1.8) $$
where the matrix $X_!$ is formed by the columns $ \widehat{X^0},\widehat X,
\dots,\widehat{X^{n-1}} $, disposed in a natural order.}
\medskip

If $X = (x_{ij}) \in {\rm Sym}(n)$, then we denote the elements of the 
matrix $X^m$ for $m = 0,1,\dots,n - 1$, by $x^{(m)}_{ij}$. 
Thus, the row of $X_!$ with number $ii$ contains the elements
$$ 1,x_{ii},x^{(2)}_{ii},\dots,x^{(n-1)}_{ii}, $$
and the line number $ ij $ with $ i < j $ is composed by the elements
$$ 
  0,\sqrt 2 x_{ij},\sqrt 2 x^{(2)}_{ij},\dots,\sqrt 2 x^{(n-1)}_{ij}.
$$

For example, in the case $n=3$ the matrix $X_!$ looks like
$$ X_! = \pmatrix {
1 & x_{11} & x^{(2)}_{11} \cr
1 & x_{22} & x^{(2)}_{22} \cr
1 & x_{33} & x^{(2)}_{33} \cr
0 & \sqrt 2 x_{12} & \sqrt 2 x^{(2)}_{12} \cr
0 & \sqrt 2 x_{13} & \sqrt 2 x^{(2)}_{13} \cr
0 & \sqrt 2 x_{23} & \sqrt 2 x^{(2)}_{23}} $$
\vskip24pt

\centerline {\bf \S 2. Discriminant polynomials}
\vskip 12pt

As we shall see below, the maximum minors of the matrix $X_!$ are closely
related to the proof of theorem 1.
We denote by $\langle i_1 j_1,\dots,i_n j_n \rangle (X)$
the minor of order $n$ of the matrix $X_!$,
formed by lines with double indices $ i_1 j_1,\dots,i_n j_n $,
This is a homogeneous polynomial of degree $C^2_n$ in the elements of
matrix $X$. For example, if $n = 3$, then
$$ \langle 11,22,33 \rangle (X) = \left \vert \matrix {
1 & x_{11} & x^2_{11} + x^2_{12} + x^2_{13} \cr
1 & x_{22} & x^2_{12} + x^2_{22} + x^2_{23} \cr
1 & x_{33} & x^2_{13} + x^2_{23} + x^2_{33}
} \right \vert = $$
$$ = (x_{22} - x_{11})(x_{33} - x_{11})(x_{33} - x_{22}) - $$
$$ - (x_{22} - x_{11})x^2_{12} + (x_{33} - x_{11})x^2_{13} -
(x_{33} - x_{22})x^2_{23}; $$
$$ \langle 33,12,23 \rangle (X) = 2 \left \vert \matrix {
x_{12} & x_{11}x_{12} + x_{12}x_{22} + x_{13}x_{23} \cr
x_{23} & x_{12}x_{13} + x_{22}x_{23} + x_{23}x_{33}
} \right \vert = $$
$$ = 2 x_{12}x_{23}(x_{33} - x_{11}) + 2 x_{13}(x^2_{12} - x^2_{23}) $$

As we already have proved, the sum of squares of all these minors is equal
to the discriminant ${\rm Dis}(\chi_X)$.
In the sequel, the minors $\langle i_1 j_1,\dots,i_n j_n \rangle (X)$ will 
be referred to as {\it discriminant polynomials}.

The central position in the set of all discriminant polynomials is occupied 
by $T(X) = \langle 11,\dots,nn\rangle(X)$.
\medskip

{\bf Proposition 4.}
{\it The polynomial
$$
   T(X) = \langle 11,\dots,nn\rangle(X)
$$
is irreducible}.
\medskip

{\sc Proof.}
The permutation group $S(n)$ acts on the set
of square matrices. Let $A = (a_{ij})$ be a matrix of order
$n$, and $\sigma\in S(n)$ a substitution. We set by definition
$(\sigma A)_{ij} = a_{\sigma (i)\sigma (j)}$.
These transformations are ring automorphisms,
in particular, for any $X \in {\rm Sym}(n)$ we have
$$\sigma (X^m) = [\sigma (X)]^m \eqno (2.1) $$

The group $S(n)$ also operates in the ring
$\{P(X)\}$ of polynomials in the variables $x_{ij}$:
$$
  \sigma P(\dots,x_{ij},\dots) = P(\dots,x_{\sigma (i)\sigma (j)},\dots)
$$
These $\sigma$'s are ring automorphisms, too.

Let us now prove the irreducibility of $T(X)$. As
$$ T(X) = \left \vert \matrix {
1 & x_{11} & x^{(2)}_{11} & \dots & x^{(n-1)}_{11} \cr
1 & x_{22} & x^{(2)}_{22} & \dots & x^{(n-1)}_{22} \cr
\vdots & \vdots & \vdots & \ddots & \vdots \cr
1 & x_{nn} & x^{(2)}_{nn} & \dots & x^{(n-1)}_{nn}
} \right \vert, \eqno (2.2)  $$
owing to (2.1) for any substitution $\sigma$ we have
$$\sigma x^{(m)}_{ij} = x^{(m)}_{\sigma (i) \sigma (j)},$$
so
$$ \sigma T(X) = \pm T(X), \eqno (2.3) $$
where the sign depends on the parity of a substitution.

We will refer to the elements $x_{ii}$ as {\it diagonal}, and to all
remaining variables as {\it nondiagonal}. It is easy to see that the element
$x^{(m)}_{ii}$ can be represented as $x^{(m)}_{ii} = x^m_{ii} + \widetilde
{x}^{(m)}_{ii}$, where each monomial of the polynomial $\widetilde
{x}^{(m)}_{ii}$ contains nondiagonal variables. Therefore, we can
write
$$T(X) = W(x_{11},x_{22},\dots,x_{nn}) + \widetilde T (X), \eqno (2.4)$$
where $W$, as before, is a Vandermonde's determinant, and each
monomial of the polynomial $\widetilde T (X)$ contains nondiagonal variables.
On the other hand, it is evident from (2.2) that every monomial in
$\widetilde T (X)$ contains diagonal variables, too.

We shall now assume that $Q(X)$ is an irreducible divisor of
$T(X)$. Due to (2.4), polynomial $Q(X)$ has the form
$$ Q(X) = R(x_{11},x_{22},\dots,x_{nn}) + \widetilde Q (X), \eqno (2.5) $$
where $R$ depends only on diagonal variables and is a divisor of $W$, and
the nondiagonal variables appear in each monomial of $\widetilde Q(X)$.
From (2.3) it follows that for any substitution $\sigma\in S(n)$ the
polynomial $\sigma(Q)$ is also an irreducible divisor of $T(X)$, and from
(2.4) and (2.5) we conclude that $\sigma(R)$ is a divisor of $W$. As
the polynomial $W$ factors into a product of linear forms $ x_{jj} -
x_{ii} $, and for any two such forms there is a substitution carrying
one of them into the other, any other irreducible divisor $T(X)$
is equal to $\sigma(Q)$ for some substitution $\sigma$.
Now we consider three cases.

1. $Q$ is a form of degree 1. Then $T(X)$ is a product of linear forms.
Also, if $\widetilde Q (X) \equiv 0$, then
$ T(X) = W(x_{11},\dots,x_{nn}) $, which obviously cannot be true. If
$\widetilde Q (X) \not\equiv 0$, then also all of $ \sigma (\widetilde Q)
\not\equiv 0$ do not depend on diagonal variables. But then $T(X)$ contains
monomials which do not depend on diagonal variables, which is also
impossible.

2. $Q$ is a form of degree greater than 1. Then $R$ is a product
of several factors $ x_{jj} - x_{ii} $, and all of
them are different. It is always possible to select such a
substitution $\sigma$, that $\sigma(R)\ne R$, but $R$ and
$ \sigma (R) $ have an identical divisor of the form $x_{jj} - x_{ii}$
(there is an exceptional case $n=4$, when the degree of $Q$ is 2;
we will discuss it separately). As the forms $Q$ and $\sigma (Q)$
are different and irreducible, they both appear in the expansion of
the polynomial $T(X)$ into irreducible factors.
Therefore $W$ is divisible by the product $R \sigma (R)$ that contains
a factor $(x_{jj} - x_{ii})^2$, and this is a contradiction.

3. The exceptional case $ n = 4 $:
$$ R = (x_{jj} - x_{ii})(x_{ll} - x_{kk}), $$
where $ (i,j,k,l) $ is a permutation of numbers 1,2,3,4. In this case $T(X)$
factors into a product of three quadratic forms. Obviously, the form
$\widetilde Q$ should have degree 1 in diagonal variables and
degree 1 in nondiagonal variables.
Then any monomial $T(X)$ has at least third degree
in diagonal variables. Let us show that this is impossible.
Assign the following values to the variables $x_{ij}$ :
$$ x_{11} = x,\qquad x_{12} = x_{23} = x_{34} = t, $$
and assume all remaining variables equal to zero. Then
$$ \langle 11,22,33,44 \rangle (X) = x^2 t^4, $$
whence $T(X)$ is obliged to contain monomials having
degree in diagonal variables no higher than 2. 
This contradiction completes the proof.
\medskip

It turns out that the discriminant polynomials can be obtained by a totally
different construction.
Let ${\rm so}(n)$ be the space of skew-symmetric matrices of order $n$.
The dimension of this space is $M = C^2_n$.
A matrix $A = (a_{kl}) \in {\rm so}(n)$ can be considered
as a column $ \bar A $ of dimension $M$, whose elements $\sqrt 2 a_{kl}$
with $k < l$ are arranged in the order (1.5), only without the first
$n$ elements. We shall consider $ {\rm so}(n) $ as a Euclidean
space of dimension $ M $ with orthonormal basis
$$ \Delta_{ij} = \frac{1}{\sqrt 2} (E_{ij} - E_{ji}) \eqno (2.6) $$
with $i<j$.

Let $ X \in {\rm Sym}(n) $. Consider the linear map
$$ X^{\Box} : {\rm so}(n) \longrightarrow {\rm Sym}(n), $$
defined by the formula
$$ X^{\Box}(A) = A X - X A $$
We denote by $X^!$ the matrix of this map in bases (1.7) and (2.6).
In the notation introduced earlier, the following
matrix-vectorial equality holds:
$ \widehat H = X^! \bar A $, where $ H = X^{\Box} (A) $. Let us describe
the matrix $X^!$. First of all it has the size $ N \times M $.
Its lines are numbered by double indices $ij$ with $ i \le j $ and are
arranged in the order (1.5). The columns are numbered by double indices
$kl$ with $ k < l $, disposed in the same order.

If
$$ \Psi_{kl} = X^{\Box} (\Delta_{kl}), $$
then $ \widehat{\Psi}_{kl}$ is the column of the matrix $ X^! $ with
number $ kl $.

Therefore the row of $ X^! $ with number $ii$ looks as follows. 
If $k<i$, then the element in the column $ki$ is $ - \sqrt 2 x_{ki}$.
If $ k > i $, the element in the column
$ik$ is $\sqrt 2 x_{ik}$. All remaining elements of this line
are zero.

Now we shall describe the line with number $ij$, where $i<j$. If $k<i$, the
element in the column $ki$ is $ -x_{kj} $, and the element in the column $
kj $ is $-x_{ki}$. If $ i < k < j $, in the column $ ik $ we have the number
$ x_{kj} $, and in the column $ kj $, the number $-x_{ik}$.
If $j<k$, the column $ik$ contains the number $ x_{jk} $, and the column
$jk$, the number $x_{ik}$. Finally, in the column $ ij $ we have the number
$x_{jj}-x_{ii}$. All remaining elements of this line are equal to zero.

For example, let $n=3$. Then
$$ X^! = \pmatrix {
\sqrt 2 x_{12} & \sqrt 2 x_{13} & 0 \cr
- \sqrt 2 x_{12} & 0 & \sqrt 2 x_{23} \cr
0 & - \sqrt 2 x_{13} & - \sqrt 2 x_{23} \cr
x_{22} - x_{11} & x_{23} & x_{13} \cr
x_{23} & x_{33} - x_{11} & -x_{12} \cr
-x_{13} & -x_{12} & x_{33} - x_{22}} $$

We are interested in the maximal minors of the matrix $ X^! $. The minor
containing all lines, except $ i_1 j_1,\dots,i_n j_n $, will be denoted
by $ \{i_1 j_1,\dots,i_n j_n \}(X) $. Obviously, it is a homogeneous
polynomial of degree $M$ in the elements $x_{ij}$ of the matrix $X$.
For example, for $ n = 3 $ we have
$$ \{11,22,33\}(X) = \left \vert \matrix {
x_{22} - x_{11} & x_{23} & x_{13} \cr
x_{23} & x_{33} - x_{11} & -x_{12} \cr
-x_{13} & -x_{12} & x_{33} - x_{22}
} \right \vert = \langle 11,22,33 \rangle (X); $$
$$ \{33,12,23\}(X) = \left \vert \matrix {
\sqrt 2 x_{12} & \sqrt 2 x_{13} & 0 \cr
- \sqrt 2 x_{12} & 0 & \sqrt 2 x_{23} \cr
x_{23} & x_{33} - x_{11} & - x_{12}
} \right \vert = -\langle 33,12,23 \rangle(X); $$

In the next section, we shall prove that a similar relation
is true in general.
\medskip

{\bf Proposition 5.}
{\it For any set of indices the following equality holds\/}:
$$ \{i_1 j_1,\dots,i_n j_n \}(X) =
\pm \langle i_1 j_1,\dots,i_n j_n \rangle (X) \eqno (2.7) $$
\vskip 24pt

\centerline {\bf \S 3. On Hodge's isomorphism for the space
of symmetric operators}
\vskip 12pt

To avoid tedious computations with associated matrices
in the proof of equality (2.7), we shall work
with linear operators rather than with matrices.
In doing so, we preserve all previous notation, using bold face
to indicate the invariant objects. We will rely upon
a well-known property of Hodge's isomorphism [3]
stating that the coordinates of skew-symmetric tensors that correspond
to each other under this isomorphism, can differ only by a sign. 
Now we introduce the necessary notations and
briefly remind some necessary properties
of invariant objects.

Let ${\cal L}$ be a real Euclidean space of dimension $n$ with
scalar product $(\ ,\ )$. By ${\bf End}(n)$ we denote the
space of linear operators over ${\cal L}$. The zero operator
will be denoted by ${\bf O}$.

An orthonormal basis ${\bf e}_1,\ldots,{\bf e}_n$ of
the space ${\cal L}$ gives rise, in a natural way,
to an orthonormal basis in the space ${\bf End}(n)$.
First we define the tensor product $({\bf a}\otimes{\bf b})$ for
an ordered pair of vectors ${\bf a},{\bf b}\in {\cal L}$. Namely,
$({\bf a}\otimes {\bf b}) \in {\bf End}(n)$ is the linear operator
defined by the formula
$$ 
  ({\bf a}\otimes {\bf b})({\bf x}) = ({\bf b},{\bf x})\skr {\bf a}.
$$

All possible products $({\bf e}_i\otimes {\bf e}_j)$ of the elements
of a basis of ${\cal L}$ constitute a basis of the space ${\bf End}(n)$, 
which we will call the accompanying basis of ${\bf e}_1,\dots,{\bf e}_n$.
Obviously, for any $i,j,k$ we have
  $$ ({\bf e}_i\otimes {\bf e}_j)({\bf e}_k) =
     \delta_{jk} {\bf e}_i, \eqno (3.1) $$
where $\delta_{jk}$ is the Kronecker delta. The operator
$({\bf e}_i\otimes{\bf e}_j)$ in the basis ${\bf e}_1,\dots,{\bf e}_n$
corresponds to the matrix $E_{ij}$. If $A = (a_{ij})$ is the
matrix of an operator ${\bf A} \in {\bf End}(n)$ in that basis, then
$$ {\bf A} = \sum_{i=1}^n\sum_{j=1}^n a_{ij} ({\bf e}_i
     \otimes {\bf e}_j). $$

In the space ${\bf End}(n)$ there is a natural scalar
product $\langle\; ,\; \rangle$, arising from the product $(\ ,\ )$
in ${\cal L}$. For any two operators $ {\bf A,B} \in {\bf End}(n) $ their
scalar product is
  $$ \langle {\bf A}, {\bf B} \rangle = {\rm tr}\skr ({\bf A^* B}) $$
where ${\bf A^*}$ is the operator conjugate to ${\bf A}$, and
${\rm tr}$ is the trace. 
The scalar products of the basic elements are:
  $$ \langle ({\bf e}_i\otimes {\bf e}_j),
     ({\bf e}_k\otimes {\bf e}_l) \rangle
     = \delta_{ik}\skr \delta_{jl}, \eqno (3.2) $$
hence the accompanying basis is orthonormal, and the coordinates 
of an operator ${\bf A} \in {\bf End}(n)$ in this basis coincide with 
the elements of its matrix in the basis ${\bf e}_1,\ldots,{\bf e}_n$.

An operator ${\bf A}$ is called {\it symmetric}, if ${\bf A^*} = {\bf A}$, 
and {\it skew-symmetric}, if ${\bf A^*}=-{\bf A}$. The subspace of symmetric
operators we denote by ${\bf Sym}(n)$, and the subspace of skew-symmetric
operators, by ${\bf so}(n)$.

Let ${\bf X} \in {\bf Sym}(n)$. Since the matrix $X = (x_{ij})$ of
${\bf X}$ in the basis ${\bf e}_1,\ldots,{\bf e}_n$ is symmetric, we have:
$$ {\bf X} = \sum_{i=1}^n x_{ii}\skr ({\bf e}_i\otimes {\bf e}_i) +
   \sum_{i<j} x_{ij}\skr (({\bf e}_i\otimes {\bf e}_j) +
   ({\bf e}_j\otimes {\bf e}_i)).
$$

The symmetric operators
  $$ {\bf \Gamma}_{ii} = ({\bf e}_i\otimes {\bf e}_i),\qquad
     {\bf \Gamma}_{ij} = \frac{1}{\sqrt{2}}\skr ({\bf e}_i\otimes
     {\bf e}_j) + \frac{1}{\sqrt{2}}\skr ({\bf e}_j\otimes
     {\bf e}_i), \eqno (3.3) $$
where $i<j$, correspond to the
matrices $ \Gamma_{ii} $ and $ \Gamma_{ij} $, introduced above. 
Any operator ${\bf X}\in{\bf Sym}(n)$ is a linear combination of the 
basic elements (3.3):
  $$ {\bf X} = \sum_{i=1}^n x_{ii}\skr {\bf \Gamma}_{ii} +
     \sqrt{2}\skr \sum_{i<j} x_{ij}\skr {\bf \Gamma}_{ij} $$
Besides, (3.2) implies that the system (3.3) is orthonormal.
Therefore, this system constitutes an orthonormal basis
of $N$-dimensional Euclidean space ${\bf Sym}(n)$, and the coordinates of
an operator $ {\bf X} \in {\bf Sym}(n) $ in this basis are the diagonal 
elements $x_{ii}$ of its matrix $X$ in the basis ${\bf e}_1,\ldots,{\bf e}_n$,
as well as nondiagonal elements $x_{ij}$ multiplied on $\sqrt{2}$.
We always put the elements of the basis (3.3) in the order (1.5).

Now let ${\bf A}\in{\bf so}(n)$. Since the matrix $ A = (a_{ij}) $ of the
operator ${\bf A}$ in the basis ${\bf e}_1,\dots,{\bf e}_n$ is skew-symmetric,
we have
  $$ {\bf A} = \sum_{i<j}\skr a_{ij}\skr (({\bf e}_i\otimes {\bf e}_j)
     - ({\bf e}_j\otimes {\bf e}_i)), $$
therefore in the space ${\bf so}(n)$ there is a natural orthonormal basis
  $$ {\bf \Delta}_{ij} = \frac{1}{\sqrt{2}}\skr ({\bf e}_i\otimes
     {\bf e}_j) - \frac{1}{\sqrt{2}}\skr ({\bf e}_j\otimes {\bf e}_i)
     \eqno (3.4) $$
with $ i < j $. We arrange the elements of this basis in the usual order
(1.5). Let $\Delta_{ij}$ be the corresponding matrices.
The coordinates of an operator ${\bf A}\in{\bf so}(n)$ in this basis
are nondiagonal elements $a_{ij}$, $i<j$, of its matrix $A$ in the
basis ${\bf e}_1,\ldots,{\bf e}_n$, multiplied by $\sqrt2$.

From now on, we fix an orientation in the space ${\cal L}$
and consider as admissible only proper orthogonal transformations
(with determinant 1).
Under such transformations in ${\cal L}$ the
accompanying bases (3.3) and (3.4) are also exposed to proper orthogonal
transformations. Indeed, suppose that two bases ${\bf e}_1,\ldots,{\bf e}_n$
and ${\bf e}'_1,\ldots,{\bf e}'_n$ are connected by the formulas
$$ {\bf e}'_i = \sum_{j=1}^n c_{ji} {\bf e}_{j}, $$
where the matrix $ C $ is proper orthogonal, $ \det C = 1 $. Then
the accompanying transformation is given by
  $$ \begin{array}{c}
     {\bf \Gamma}'_{ii} = \sum\limits_{k=1}^n c^2_{ki}\skr
     {\bf \Gamma}_{kk} + \sqrt{2}\skr\sum\limits_{k<l} c_{ki}\skr
     c_{li}\skr {\bf \Gamma}_{kl} \\
                  \\
     {\bf \Gamma}'_{ij} = \sqrt{2}\skr\sum\limits_{k=1}^n c_{ki}\skr
     c_{kj}\skr {\bf \Gamma}_{kk} + \sum\limits_{k<l} (c_{ki}\skr
     c_{lj} + c_{kj}\skr c_{li})\skr {\bf \Gamma}_{kl} \\
                   \\
     {\bf \Delta}'_{ij} = \sum\limits_{k<l} (c_{ki}\skr c_{lj}
     - c_{kj}\skr c_{li})\skr {\bf \Delta}_{kl}
     \end{array} \eqno (3.5) $$

Fort an operator in ${\cal L}$ we denote by $C^{\wedge}$ and 
$C_{\wedge}$ the matrices of accompanying transformations in
the spaces ${\bf Sym}(n)$ and ${\bf so}(n)$, respectively.
Since all bases connected by formulas (3.5),
are orthonormal, both matrices $ C^{\wedge} $ and $ C_{\wedge} $ are
orthogonal. For example, if $n=2$ and
$$ C = \left(\matrix{
     a & -b \cr b & a \cr}\right) ,
    \qquad\mbox{where}\qquad a^2 + b^2 = 1,
$$
then
$$
     C^{\wedge} = \left(\matrix{
     a^2 & b^2 & -\sqrt{2}\skr a\skr b \cr
     b^2 & a^2 & \sqrt{2}\skr a\skr b \cr
     \sqrt{2}\skr a\skr b & -\sqrt{2}\skr a\skr b & a^2 - b^2 \cr}
     \right) \qquad\mbox{and}\qquad C_{\wedge} = (1).
$$

Now let us prove that the matrices $C^{\wedge}$ and $C_{\wedge}$ are proper
orthogonal, that is their determinants are positive. Indeed, since the matrix
$C$ is proper orthogonal, it can be continuously transformed
to the unit matrix. More exactly, there are such continuous functions 
$c_{ij}(t)$, defined for $0\le t \le 1$, that $c_{ij}(0) = \delta_{ij}$,
$c_{ij}(1) = c_{ij} $, and the matrix $ C(t) = (c_{ij}(t)) $ is orthogonal
for all values of $ t $. Considering the family of transformations (3.5),
where the elements $c_{ij}$ are replaced by $ c_{ij}(t) $, we
obtain a continuous set of orthonormal bases $ {\bf \Gamma}'_{ij}(t) $
and $ {\bf \Delta}'_{ij}(t) $ that satisfy
$$ {\bf \Gamma}'_{ij}(0) = {\bf \Gamma}_{ij},\quad
   {\bf \Gamma}'_{ij}(1) = {\bf \Gamma}'_{ij},\qquad
   {\bf \Delta}'_{ij}(0) = {\bf \Delta}_{ij},\quad
   {\bf \Delta}'_{ij}(1) = {\bf \Delta}'_{ij} $$
Therefore, matrices $ C^{\wedge} $ and $ C_{\wedge} $ can be continuously
deformed to unit matrices of appropriate dimensions and thus have
positive determinant.

We have proved that an orientation in space ${\cal L}$ naturally induces 
an orientation in spaces ${\bf Sym}(n)$ and ${\bf so}(n)$.  Henceforth, 
we will assume the bases (3.3) and (3.4) to be positively oriented.

The dimensions of spaces $ {\bf Sym}(n) $ and $ {\bf so}(n) $ are connected
by the relation $ M + n = N $.
We shall study the exterior algebra
over the space ${\bf Sym}(n)$:
  $$ \bigwedge\skr ({\bf Sym}(n)) = \bigoplus\limits_{p=0}^N
     \bigwedge\nolimits^p\skr ({\bf Sym}(n)) \eqno (3.6) $$

In this algebra we are interested in two multivectors $ {\bf X_!} $ and
$ {\bf X^!} $ of degrees $ n $ and $ M $ respectively, associated with an 
operator ${\bf X}\in{\bf Sym}(n)$. The first one is defined as
the exterior product of powers $ {\bf X}^q $ of the operator $ {\bf X} $ for
$ q = 0,1,\ldots,n - 1 $:
  $$ {\bf X_!} = {\bf X}^0\wedge {\bf X}^1\wedge\cdots
     \wedge {\bf X}^{n-1} $$

The definition of the second multivector is a little bit more involved.
Let
  $$ {\bf X^{\Box}}: {\bf so}(n)\longrightarrow {\bf Sym}(n), $$
be the linear map given by
  $$ {\bf X^{\Box}}\skr ({\bf A}) = {\bf A\skr X} - {\bf X\skr A} $$
Let $ {\bf \Phi}_1,\ldots,{\bf \Phi}_M $ be an arbitrary positively
oriented orthonormal basis of space $ {\bf so}(n) $. Denote
$ {\bf \Psi}_r = {\bf X^{\Box}}\skr ({\bf \Phi}_r) $. We set by 
definition
  $$ {\bf X^!} = {\bf \Psi}_1\wedge {\bf \Psi}_2
     \wedge\cdots\wedge {\bf \Psi}_M $$

To prove the correctness of this definition, suppose that $ {\bf \Phi}'_1,
\ldots,{\bf \Phi}'_M $ is another positively oriented orthonormal
basis of space $ {\bf so}(n) $, connected with the first basis by a proper
orthogonal matrix $G$. By the linearity of operator $ {\bf X^{\Box}} $, 
systems of vectors $ {\bf \Psi}_1,\ldots,
{\bf \Psi}_M $ and $ {\bf \Psi}'_1,\ldots,{\bf \Psi}'_M $, where
$ {\bf \Psi}'_r = {\bf X^{\Box}}\skr ({\bf \Phi}'_r) $, are related by the
same matrix $ G $. And since $ \det G = 1 $, we have
  $$ {\bf \Psi}'_1\wedge\cdots\wedge {\bf \Psi}'_M = (\det G)\skr
     {\bf \Psi}_1\wedge\cdots\wedge {\bf \Psi}_M = {\bf X^!} $$

Matrices $ X_! $ and $ X^! $ considered above are
matrices of multivectors $ {\bf X_!} $ and $ {\bf X^!} $, if $ X $ is
the matrix of the operator $ {\bf X} $, and 
operators $ {\bf \Delta}_{ij} $ are used as basis vectors $ {\bf \Phi}_r $
in $ {\bf so}(n) $.
The minors $ \langle i_1 j_1,\dots,i_n j_n \rangle(X) $ and $ \{i_1 j_1,\dots,
i_n j_n \}(X) $ represent the coordinates of multivectors $ {\bf X_!} $ and
$ {\bf X^!} $, respectively.

There is a very simple connection between the multivectors 
$ {\bf X_!} $ and $ {\bf X^!} $.
To find it, we note that the summands of complementary degrees in
formula (3.6) are related by the Hodge isomorphism
$$ {\cal H}_p: \bigwedge\nolimits^p\skr ({\bf Sym}(n)) \longrightarrow
     \bigwedge\nolimits^{N-p}\skr ({\bf Sym}(n)).
$$
In particular, the spaces of skew-symmetric tensors of degrees $ n $
and $ M $ are isomorphic, because $M=N-p$.
With the orientation of ${\bf Sym}(n)$ that we have chosen
the following relation holds:
  $$ \begin{array}{c}
     {\cal H}_n\skr ({\bf \Gamma}_{11}\wedge {\bf \Gamma}_{22}
     \wedge\cdots\wedge {\bf \Gamma}_{nn}) = \\
              \\
     = {\bf \Gamma}_{12}\wedge\cdots\wedge {\bf \Gamma}_{1n}\wedge
     {\bf \Gamma}_{23}\wedge\cdots\wedge {\bf \Gamma}_{2n}\wedge
     \cdots\wedge {\bf \Gamma}_{n-1,n} \end{array} \eqno (3.7) $$

We shall prove that the Hodge isomorphism $ {\cal H}_n $ transforms
the multivector $ {\bf X_!} $ into the multivector $ {\bf X^!} $, that is
  $$ {\cal H}_n\skr ({\bf X_!}) = {\bf X^!} \eqno (3.8) $$

First of all, relations (3.1), (3.3) and (3.4) imply that
  $$ {\bf \Gamma}^{\Box}_{ii}\skr ({\bf \Delta}_{ij}) =
     -{\bf \Gamma}_{ij},\quad {\bf \Gamma}^{\Box}_{jj}\skr
     ({\bf \Delta}_{ij}) = {\bf \Gamma}_{ij},\quad
     {\bf \Gamma}^{\Box}_{kk}\skr ({\bf \Delta}_{ij}) =
     {\bf O}, \eqno (3.9) $$
if $i<j$ and $k\ne i,j$.

Further, we can choose a basis ${\bf e}_1,\ldots,{\bf e}_n$ of ${\cal L}$
in which the matrix of ${\bf X}$ is diagonal:
  $$ X = {\rm diag}(\lambda_1,\dots,\lambda_n) $$
Then for any $q\ge0$
  $$ {\bf X}^q = \lambda_1^q\skr {\bf \Gamma}_{11} + \lambda_2^q\skr
     {\bf \Gamma}_{22} + \cdots + \lambda_n^q\skr {\bf \Gamma}_{nn}, $$
and, therefore,
  $$ {\bf X_!} = W\skr (\lambda_1,\ldots,\lambda_n)\skr {\bf \Gamma}_{11}
     \wedge {\bf \Gamma}_{22}\wedge\cdots\wedge {\bf \Gamma}_{nn} $$

Finally, from (3.9) it follows that
  $$ {\bf X^{\Box}}\skr ({\bf \Delta}_{ij}) =
     (\lambda_j - \lambda_i)\skr {\bf \Gamma}_{ij}, $$
whence
  $$ {\bf X^!} = \prod\limits_{i<j} (\lambda_j - \lambda_i)\times
     {\bf \Gamma}_{12}\wedge\cdots\wedge {\bf \Gamma}_{1n}\wedge
     {\bf \Gamma}_{23}\wedge\cdots\wedge {\bf \Gamma}_{2n}
     \wedge\cdots\wedge {\bf \Gamma}_{n-1,n}.$$
Here the coefficient of the exterior product is the Vandermonde
determinant $W(\lambda_1,\ldots,\lambda_n)$. The equality (3.8) now follows
from (3.7).

In an arbitrary coordinate system multivectors ${\bf X_!}$ and ${\bf X^!}$
have representations
  $$ {\bf X_!} = \sum \langle i_1 j_1,\dots,i_n j_n \rangle (X)\skr
     {\bf \Gamma}_{i_1 j_1}\wedge\cdots\wedge {\bf \Gamma}_{i_n j_n} $$
  $$ {\bf X^!} = \sum\{i_1 j_1,\dots,i_n j_n\}(X)\skr {\bf \Gamma}_{k_1 l_1}
     \wedge\cdots\wedge {\bf \Gamma}_{k_M l_M}, $$
where $ k_1 l_1,\dots,k_M l_M $ is the set of double indexes complementary
to the set $ i_1 j_1,\dots,i_n j_n $ in the sequence (1.5). By Hodge's
isomorphism we have
  $$ {\cal H}_n ({\bf \Gamma}_{i_1 j_1}\wedge\cdots\wedge {\bf \Gamma}
_{i_n j_n}) = \pm\skr {\bf \Gamma}_{k_1 l_1}
\wedge\cdots\wedge {\bf \Gamma}_{k_M l_M}, $$
and equality (2.7) follows from (3.8). Due to (2.7), the minors $ \{i_1 j_1,
\dots,i_n j_n \}(X) $ can also be called discriminant polynomials.
\vskip 24pt

\centerline {\bf \S 4. Proof of the main theorems}
\vskip 12pt

{\sc Proof of Theorem 1.}
A matrix $X\in{\rm Sym}(n)$
will be called {\it regular}, if all its eigenvalues are distinct.
The set of all regular matrices will be denoted by $\mathfrak{R}$.
Matrices belonging to ${\rm Sym}(n)\setminus\mathfrak{R}$ will be
referred to as {\it critical}.

Let $X$ be a regular matrix. Then there is an orthogonal matrix
$C$ such that
  $$ C^{-1} X C = L = {\rm diag} (\lambda_1,
     \lambda_2,\dots,\lambda_n), \eqno (4.1) $$
and all numbers $ \lambda_1,\lambda_2,\dots,\lambda_n $ are pairwise 
different.
By our assumptions
  $$ C^{-1} {\cal F} (X) C = M = {\rm diag} (\mu_1,\mu_2,\dots,\mu_n) $$
Diagonal matrices $E,L,\dots,L^{n-1}$ are linearly independent, therefore
diagonal matrix $ M $ can be (uniquely) represented as their linear 
combination:
  $$ M = a_0 E + a_1 L + \cdots + a_{n-1} L^{n-1}, $$
whence
  $$ {\cal F}(X) = a_0 E + a_1 X + \cdots + a_{n-1} X^{n-1}. $$
If the matrix $X$ is regular, then the coefficients $a_0,a_1,\dots,a_{n-1}$
are uniquely determined by the elements of $X$.

In what follows, by smoothness of a function we understand its infinite
smoothness. Let us prove that functions $ a_0 (X),a_1 (X),\dots,
a_{n-1}(X) $ are smooth on the set $ \mathfrak{R} $. Denote the elements of
the matrix ${\cal F}(X)$ by $\theta_{ij}(X)$.
For any matrix $X\in\mathfrak{R}$ and any collection of double indices
$ i_1 j_1,i_2 j_2,\dots,i_n j_n $ the equality
$$ \pmatrix {
x^{(0)}_{i_1 j_1} & x^{(1)}_{i_1 j_1} & \dots & x^{(n-1)}_{i_1 j_1} \cr
x^{(0)}_{i_2 j_2} & x^{(1)}_{i_2 j_2} & \dots & x^{(n-1)}_{i_2 j_2} \cr
\vdots & \vdots & \ddots & \vdots \cr
x^{(0)}_{i_n j_n} & x^{(1)}_{i_n j_n} & \dots & x^{(n-1)}_{i_n j_n}}
\pmatrix {
a_0 (X) \cr a_1 (X) \cr \vdots \cr a_{n-1}(X)} =
\pmatrix {
\theta_{i_1 j_1}(X) \cr \theta_{i_2 j_2}(X) \cr
\vdots \cr \theta_{i_n j_n}(X)}, $$
is fulfilled.
The right hand member of this equality is defined and smooth on all of
${\rm Sym}(n)$.
If $i_r<j_r$, we multiply line number $r$ by $\sqrt2$.
Multiplying after that both parts of the equality on the left by the matrix
adjoined to the matrix in the left hand side, we obtain,
for all $m=0,1,\dots,n-1$, the following relations:
  $$ \langle i_1 j_1,\dots,i_n j_n \rangle (X) \skr a_m (X) =
     \varphi_{i_1 j_1,\dots,i_n j_n,m}(X), \eqno (4.2) $$
where all functions $\varphi_{i_1 j_1,\dots,i_n j_n,m}$ are defined and
smooth on the entire space $ {\rm Sym}(n) $.

Let $X_0\in\mathfrak{R}$. In the point $X_0$, matrices $E,X,\dots,X^{n-1}$
are linearly independent, that is the rank of matrix $ X_! $ is exactly
$n$. Therefore, $X_!$ contains a minor
$\langle i_1 j_1,\dots,i_n j_n \rangle(X)$, nondegenerate in the point $X_0$.
There is a neighbourhood $\Omega\ni X_0$, such that everywhere in $\Omega$
we have
  $$ \langle i_1 j_1,\dots,i_n j_n \rangle (X) \ne 0 $$
This, together with (4.2), implies that $a_m(X)\in C^{\infty}(\Omega)$ for
all $m$.

Functions $a_m(X)$ smoothly extend to the critical set.
Indeed, as we saw before, for any $m = 0,1,\dots,n-1$ and any collection
of double indices $k_1 l_1,\dots,k_n l_n$ the function
$f(X)$, defined on the set $\mathfrak{R}$ by the equality
  $$ f(X) = \langle k_1 l_1,\dots,k_n l_n \rangle (X)\skr a_m (X), $$
can be smoothly prolonged to the entire space ${\rm Sym}(n)$. Therefore, for
any linear combination of the minors $\langle k_1 l_1,\dots,k_n l_n\rangle(X)$
  $$ P(X) = \sum P_{k_1 l_1,\dots,k_n l_n}
     \langle k_1 l_1,\dots,k_n l_n \rangle (X), $$
where all $P_{k_1 l_1,\dots,k_n l_n} $ are constants, the function
$g(X)=P(X)\skr a_m(X)$ is smooth on ${\rm Sym}(n)$. Let
$Q(X)$ be another linear combination and $h(X) = Q(X)\skr a_m(X)$. 
Obviously,
  $$ h(X)\skr P(X) = g(X)\skr Q(X) \eqno (4.3) $$
in all points of $\mathfrak{R}$. As the critical matrices make in
${\rm Sym}(n)$ a set of measure zero and functions $g(X)$,
$h(X)$ are smooth everywhere, equality (4.3) holds on the whole set
${\rm Sym}(n)$. Now, if we could prove, proceeding from (4.3),
that $g(X)$ is divisible by $P(X)$ in a neighbourhood of a critical point
$X_0$ and the quotient is a smooth function, then we would be able
to smoothly prolong the function $a_m(X)$ to a neighbourhood of $X_0$.
To establish this fact, under a special choice of polynomials $P(X)$ and
$Q(X)$, we will need the Malgrange theorem [4].
\smallskip

{\sc The Malgrange theorem.} {\it Let $\Omega$ be an open set
in $R^N$ and $f_1,\dots,f_r$ be analytical functions
in $\Omega$. Let $\varphi\in C^{\infty}(\Omega)$. Then
$\varphi$ is representable as
  $$ \varphi = \sum_{i=1}^r f_i \psi_i, $$
where $\psi_i\in C^{\infty}(\Omega)$, if and only if
for every $a\in\Omega$ the Taylor expansion $T_a\varphi$ belongs to the
ideal generated by the expansions
$T_a f_i$ in the ring of formal power series in the point $a$}.
\smallskip

The Malgrange theorem is about the division of
smooth functions by analytical functions.
In the present investigation, analytical functions will always be 
polynomials.
Therefore we shall touch upon the problem of division
of formal power series by a polynomial. 
A polynomial of several variables can be uniquely represented as a sum of
homogeneous polynomials. The one that has the smallest degree, is usually
called the {\it initial form\/} of the given polynomial.

We shall use the following sufficient condition of divisibility. Let $ P_1 $
and $ P_2 $ be polynomials in several variables, while $ R_1 $ and $ R_2 $
be formal power series in the same variables, such that
$ R_1 P_2 = R_2 P_1 $. Suppose that the initial forms of polynomials
$ P_1 $ and $ P_2 $ are coprime. It is easy to prove that then the
polynomials $P_1$ and $P_2$ must be coprime in the ring of formal power 
series. Since this ring is factorial, the series $ R_1 $ is divisible by
$P_1$, and the series $R_2$ by $P_2$.

Here we encounter the problem about the degrees of the 
initial forms of discriminant polynomials, decomposed into a finite 
Taylor series in a neighbourhood of a critical point $X_0$. 
Formula (1.8) suggests that for the solution of this problem it is
important to find the multiplicity of a zero of the discriminant 
${\rm Dis}(\chi_X)$ in a critical point.

Below, we will use the following notion.
An integer sequence $(m_1,m_2,\dots,m_k)$ is the {\it root type\/} 
of a matrix $X\in{\rm Sym}(n)$, if $X$ has exactly $k$ different
eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_k$, and $m_i$ is the
multiplicity of $\lambda_i$. Of course, the sum $m_1+m_2+\cdots+m_k$ is 
always equal to $n$. We say that two root types $(m_1,m_2,\dots,m_k)$ and
$(r_1,r_2,\dots,r_l)$ are equivalent, if $k=l$ and the two sequences differ
only by a permutation.

Further, we will say that the root type $(r_1,r_2,\dots,r_l)$ is obtained from
the root type $(m_1,m_2,\dots,m_k)$ by an elementary operation, if
$l=k-1$ and there are integers $\alpha,\beta$: $1\le\alpha<\beta\le k$, 
that the root type
  $$ (m_1,\dots,m_{\alpha - 1},m_{\alpha} + m_{\beta},m_{\alpha + 1},
     \dots,m_{\beta - 1},m_{\beta + 1},\dots,m_k) $$
is equivalent to $(r_1,r_2,\dots,r_l)$. 
Finally, a root type $A$ precedes a root type $B$, if $B$ can be obtained
from $A$ by means of several elementary operations.

Evidently, the root type $(1,1,\dots,1)$ of a regular matrix
precedes any other, and any root type precedes $(n)$, the root type
of a scalar matrix.

Let $X_0\in{\rm Sym}(n)$ be a critical matrix of
root type $(m_1,m_2,\dots,m_k)$. We are interested
in the behaviour of linear combinations of discriminant polynomials in the
neighbourhood of $X_0$ and the multiplicity of the zero of the discriminant
${\rm Dis}(\chi_X)$ in the point $X_0$.
The matrix $X_0$ can be reduced to diagonal form by a conjugation
$Y_0=C X_0 C^{-1}$.
Transformation (1.6) gives a one-to-one correspondence between the
neighbourhoods of points $X_0$ and $Y_0$. Viewing ${\rm Sym}(n)$ as a
Euclidean space of dimension $N$ this transformation can be written as
$\widehat X = C^{\wedge}\widehat Y$, where $C^{\wedge}$ is an
orthogonal matrix of order $N$. Equality (1.6) implies that for any
$q$ we have $X^q = C^{-1} Y^q C$, or in another notation,
$\widehat{X^q}=C^{\wedge}\widehat{Y^q}$.
Setting $q=0,1,\dots,n-1$, we get the transformation rule
  $$ X_! = C^{\wedge} Y_! $$

This, together with the Binet--Cauchy formula, shows that polynomials 
$\langle i_1 j_1,\dots,i_n j_n \rangle(X)$ are constant coefficient linear
combinations of polynomials $\langle i_1 j_1,\dots,i_n j_n\rangle(Y)$
and vice versa. As the discriminant is invariant with respect to
similarity transformations, we can assume that
$X_0$ is a diagonal matrix
  $$ X_0 = \left(\matrix{
     L_1 &&& \cr & L_2 && \cr && \ddots & \cr &&& L_k}\right),
     \qquad\mbox{where}\qquad
     L_{\alpha} = \left(\matrix{
     \lambda_{\alpha} && \cr
     & \ddots & \cr
     && \lambda_{\alpha}}
     \right) \eqno (4.4) $$
is a scalar matrix of dimension $ m_{\alpha} $. Now it is easy to prove
(using, for example, Gershgorin's circles [2]), that the zero
multiplicity in question is equal to $2\nu$, where
  $$ \nu = C^2_{m_1} + C^2_{m_2} + \dots + C^2_{m_k} \eqno(4.5)$$
Taking into account (1.8), we conclude that any discriminant polynomial
has, in the point $X_0$, a zero of order greater than or equal to $\nu$.

Return to the proof of the theorem. We have estimated from below the degrees
of the initial forms of discriminant polynomials expanded in the Taylor
series in a neighbourhood of any critical point. Now we can construct
polynomials $P(X)$ and $Q(X)$ with relatively prime initial forms for
equality (4.3).

As we already did earlier, we shall divide the set of variables $x_{ij}$
into two groups: diagonal variables $ x_{ii} $ and nondiagonal variables
$ x_{ij} $ with $ i < j $. Suppose that $ R(Z,W) $ is
a homogeneous polynomial of degree $M$ depending on two groups of variables
$ Z = (z_1,z_2,\dots,z_k) $ and $ W = (w_1,w_2,\dots,w_l) $. Then it can be
uniquely represented as
  $$ R(Z,W) = \sum_{s+t=M} R_{s,t} (Z,W), $$
where each polynomial $ R_{s,t} (Z,W) $ is homogeneous of degree $s$
in variables $Z$ and homogeneous of degree $t$ in variables $W$.
The nonzero summand $R_{s,t}(Z,W)$ with the greatest $s$
will be referred to as the {\it highest part of $R(Z,W)$ with respect to
variables $Z$}. If two homogeneous polynomials $R_1(Z,W)$ and $R_2(Z,W)$
have coprime highest parts with respect to $Z$, then the polynomials
themselves are coprime, too.

Denote by $\Delta K(X)$ the highest part of a homogeneous polynomial $K(X)$
with respect to diagonal variables. For any $m$, elements $x^{(m)}_{ij}$ of
the matrix $X^m$ are homogeneous polynomials of degree $m$. It is easy
to see that
  $$ \Delta x^{(m)}_{ii} = x^m_{ii},\qquad \Delta x^{(m)}_{ij} =
     x_{ij}\skr \frac{x^m_{jj} - x^m_{ii}}{x_{jj} - x_{ii}} \eqno (4.6) $$
This implies the formula
  $$ \Delta \langle 11,22,\dots,nn \rangle (X) =
     W(x_{11},x_{22},\dots,x_{nn}), \eqno (4.7) $$
where $W$ is the Vandermonde determinant. Further,
let $1\le i<j\le n$ and denote by $P_{ij}(X)$ the determinant obtained 
from the determinant $\langle 11,22,\dots,nn\rangle(X)$, if its row $jj$ 
is replaced by the row $ij$. Due to (4.6) we have
  $$ \Delta P_{ij} (X) = \frac{x_{ij}}{x_{jj} - x_{ii}}\skr
     W(x_{11},x_{22},\dots,x_{nn}) \eqno (4.8) $$

Now let $X_0$ be a critical matrix. Let's prove, that the coefficients
$a_m(X)$ can be continued to a neighbourhood of $X_0$ with smoothness
preserved. The proof proceeds by induction on the root type of $X_0$.
The induction base is the root type $ (1,1,\dots,1) $ of regular matrices.
Suppose that such prolongation is possible for all points,
whose root type precedes that of $X_0$.

We can assume that $X_0$ is a diagonal matrix (4.4). 
Choose a neighbourhood $\Omega$
of the point $X_0$ such that the root type of all its points
precedes or is equivalent to the type of $X_0$, and set $X=X_0+H$.
The initial form in the Taylor expansion of the polynomial $K(X)$ in the point
$X_0$ will be denoted by $K_{X_0}(H)$. Further, we will say that a pair of
indices $(i,j)$ is admissible, if there exists an integer $t$ such that
  $$ m_0 + m_1 + \cdots + m_t + 1 \le i < j \le
     m_0 + m_1 + \cdots + m_{t+1} $$
(we assume that $m_0=0$, $0\le t\le n-1$). Setting
  $$ Q(X) = \langle 11,22,\dots,nn \rangle (X) $$
in equality (4.3), we can see that $Q(X)$ has in the point $X_0$
a zero of order $\nu$, defined by (4.5).
By (4.7), the highest part of the initial form $Q_{X_0}(H)$ 
with respect to diagonal variables is
  $$ \Delta Q_{X_0}(H) = B\skr \prod (h_{jj} - h_{ii}), $$
where $B\ne0$ is a constant depending only on $X_0$, and the
product is taken over all admissible pairs.

In the same way, if $(i,j)$ is an admissible pair, then by (4.8)
the polynomial $P_{ij}(X)$ in a vicinity of $X_0$ satisfies
  $$ \Delta P_{ij,X_0} (H) = \frac{h_{ij}}{h_{jj} - h_{ii}}\skr
     \Delta Q_{X_0} (H). $$

Therefore, if we put, in equality (4.3),
  $$ P(X) = \sum P_{ij}(X), $$
with the summation going over all admissible pairs, we will obtain
  $$ \Delta P_{X_0}(H) = \Delta Q_{X_0}(H)
     \sum \frac{h_{ij}}{h_{jj} - h_{ii}}, $$
which shows that polynomials $\Delta P_{X_0}(H)$ and
$\Delta Q_{X_0}(H)$ are coprime. Then the initial forms $P_{X_0}(H)$
and $Q_{X_0}(H)$ are coprime, too, and in the point $X_0$ the formal series
$T_{X_0}g(X)$ is divisible by $T_{X_0}P(X)$.

Furthermore, in all points of the neighbourhood $\Omega$, whose root type
precedes the root type of $X_0$, by the inductive assumption the coefficient
$a_m(X)$ is smooth function and hence the formal series for
$g(X)$ is divisible by the expansion of $P(X)$.

Finally, we have to consider those points in the neighbourhood $\Omega$
whose root type is equivalent to that of $X_0$. In these points
the polynomials $P(X)$ and $Q(X)$ have zero of order no smaller than $\nu$.
If the neighbourhood $\Omega$ is small enough, the forms of
degree $\nu$ of polynomials $P(X)$ and $Q(X)$ expanded in any point
of $\Omega$, will be arbitrarily close to the corresponding forms of these
polynomials expanded in the point $X_0$. Therefore, first, $P(X)$ and
$Q(X)$ in the points of $\Omega$ of equivalent root type have zero of
order exactly $\nu$, and second, the initial forms of the expansions of
$P(X)$ and $Q(X)$ in these points are coprime. Therefore in these points 
the formal expansion of $g(X)$ is divisible by the expansion of $P(X)$.

The statement that close enough forms are coprime, requires some clarification.
Let $\alpha_X(H)$ and $\beta_X(H)$ be two forms of degree $\nu$
in variables $H$ whose coefficients are continuous functions in variables
$X$. If for some value of $X$ these forms were not relatively prime, then
we would be able to find nonzero forms $\gamma(H)$ and $\delta(H)$ of 
degree $\nu-1$, such that
  $$ \alpha_X (H)\delta (H) = \beta_X (H)\gamma (H) \eqno (4.9) $$

The last equality is a system of linear equations on the coefficients
of the forms $\gamma(H)$ and $\delta(H)$. For $X=X_0$ this system has only
trivial solution, that is its rank is not less than the number of unknowns. 
The matrix of system (4.9) continuously depends on $X$. Therefore for
$X$, close enough to $X_0$, its rank is also not less than the number of
unknowns. It follows that system (4.9) has only a trivial solution, and the
forms $\alpha_X (H)$ and $\beta_X(H)$ are coprime.

So, in any point $X$ of a certain neighbourhood $\Omega\ni X_0$ the formal
series $T_X g$ belongs to the ideal generated by the series $T_X P$.
By the Malgrange theorem the function $g(X)$ is divisible by $P(X)$
in the domain $\Omega$, and the quotient is a smooth function.
We have thus proved that any coefficient $a_m(X)$ can be extended to a 
neighbourhood of the point $X_0$ as a smooth function.
As $X_0$ is an arbitrary critical point,
the functions $a_m(X)$ prolong to the entire critical set. This completes
the proof of theorem 1.
\medskip

Of course, Theorem 1 remains valid in the case, when we have
a function ${\cal F}$ defined only on a subset of the space ${\rm Sym}(n)$.
\medskip

{\sc Proof of Theorem 2.}
Theorem 2 is proved in a similar way, using equality (2.7).
The only essential difference is in the proof of the induction base,
which we now shall fulfill. Let $X$ be a regular matrix. Then there is
an orthogonal matrix $C$, such that equality (4.1) holds. Let
  $$ C^{-1} {\cal G}(X) C = T = (t_{ij}), $$
where all $t_{ii}=0$. We can choose a skew-symmetric matrix $B$, such that
  $$ T = B\skr L - L\skr B $$
Indeed, the elements $b_{ij}$ of $B$ are uniquely determined
by relations $ t_{ij} = (\lambda_j - \lambda_i)\skr b_{ij} $.

Applying an appropriate similarity transformation, we obtain
(0.2), where $A(X)=CBC^{-1}$. Thus, on the regular set $\mathfrak{R}$
the matrix $A(X)$ exists and is unique. Relation (0.2) can be written in
matrix--vectorial form
  $$ \widehat{{\cal G}(X)} = X^!\skr \overline{A(X)}, $$
and we see that the prolongation
of the matrix elements $a_{ij}(X)$ of $A(X)$ to the
critical set can be constructed in the same way as above,
using the discriminant polynomials $ \{i_1 j_1,\dots,i_n j_n \}(X) $.
\bigskip

{\bf Acknowledgments.}
The author is grateful to his teacher A.\,S.\, Mischenko for the
statement of the problems and to P.\,P.\,Mosolov,
A.\,T.\,Fomenko, S.\,V.\,Duzhin and V.\,I.\,Arnold for their interest
in this work.

\vskip 60pt
\centerline {References}
\vskip 20pt

\noindent [1] {\it A. I. Lourie.} Teorija uprugosti. Moscow, Nauka, 1970.

\noindent [2] {\it F. R. Gantmacher.} 
The Theory of Matrices Vol. 1, 2, Chelsea, New York. 1959.

\noindent [3] {\it M. M. Postnikov.} Lectures on geometry. Semester II.
Linear algebra. M.: Nauka. 1986.

\noindent [4] {\it B. Malgrange.} Ideals of differentiable functions.
London: Oxford University Press, 1966.

\noindent [5] {\it N. V. Ilyushechkin.} On one class of smooth
matrix valued functions. Uspehi Mat. Nauk. 1985. v. 40. No.~1.

\noindent [6] {\it N. V. Ilyushechkin.} Discriminant of the characteristic
polynomial of a normal matrix. Mathematical Notes. 1992. v. 51. No.~3.
p. 230 -- 235.

\noindent [7] {\it N. V. Ilyushechkin.} On certain classes of smooth
transformations of the space of symmetric matrices.
Matem. Sbornik, 1993, v. 184, No.~9.

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