Matrix Vieta Theorem

Dmitry FUCHS and Albert SCHWARZ* Department of Mathematics University of California Davis Ca 95616, USA

1. Introduction

Consider a matrix algebraic equation

X^{n} + A_{1}X^{n−1} + … + A_{n} = 0,  (1)

where the coefficients A_{1}, …, A_{n} as well as solutions X are supposed to be square complex matrices of some order k. For an usual algebraic equation of degree n the classical Vieta formulas express the coefficients in terms of the n solutions. However, a matrix degree n

equation generically has \binom{nk}{k} rather than n solutions. (Throughout this article the word

generic refers to a Zariski open set.)

We call n solutions X_{1}, …, X_{n} of the equation (1) independent if they determine the coefficients A_{1}, …, A_{n} (a more technical explanation of independence see in Section 2). However, the expressions of A_{1}, …, A_{n} in terms of X_{1}, …, X_{n} are much less elegant, than Vieta formulas (see, e.g. formulas (3) – (5) below). Still there are some relations between A’s and X’s which are very similar to Vieta.

Theorem 1.1. If solutions X_{1}, …, X_{n} of the equation (1) are independent, then

tr A_{1} = −(tr X_{1} + … + tr X_{n}),
det A_{n} = (−1)^{nk}det X_{1} … det X_{n}.

(2)

Theorem 1.1 is proved in Section 4 (a more direct proof for n = 2 is given in Section 3). In Section 5 we discuss a generalization of Theorem 1.1 from complex matrix algebras to arbitrary associative unitary rings.

Theorem 1.1 is elementary, however it is connected with some constructions of modern Mathematics. For every associative algebra A one can construct a linear space F (A) = A^{2}/[A, A]. This space appeared in [K] as the space of 0- forms on a non-commutative formal manifold. (Such a manifold is determined by a free associative algebra A. For the free associative algebra A with generators a_{1}, …, a_{n} the space F (A) is spanned by cyclic words of length ≥ 2 with letters a_{1}, … a_{n}.) The space F (A) appeared also in [GS] and [AW] in relation to other problems. It is easy to understand that the formula for tr A_{1} can be interpreted as a non-trivial identity in F (A).

* This work was partially supported by NSF grant DMS-9201366. Research at MSRI was supported by NSF grant DMS-9022140.

2. Independent matrices.

Definition 2.1. Matrices X_{1}, …, X_{n} are called independent if the bloc Vandermonde determinant is not zero:

| \begin{matrix}I & I & … & I \\ X_{1} & X_{2} & … & X_{n} \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ X_{1}^{n−1} & X_{2}^{n−1} & … & X_{n}^{n−1} \end{matrix} | ≠ 0.

For n = 2 the independence condition means that det (X_{1} − X_{2}) ≠ 0. For n ≥ 3 it does not imply and is not implied by the condition det (X_{i} − X_{j}) ≠ 0, 1 ≤ i < j ≤ n.

It is obvious that a generic equation (1) has n independent solutions (otherwise the above determinant would have vanished for any n solutions of any equation (1)). It is clear also that the matrices X_{1}, …, X_{n} are independent if and only is there exist unique A_{1}, …, A_{n} such that X_{1}, …, X_{n} satisfy the equation (1). In other words, for independent X_{1}, …, X_{n} the matrices A_{1}, …, A_{n} may be expressed via X_{1}, …, X_{n}. For example, if n = 2 then

\begin{aligned}A_{1} & = −(X_{1}^{2} − X_{2}^{2})(X_{1} − X_{2})^{−1}, \\ A_{2} & = −X_{1}^{2} + (X_{1}^{2} − X_{2}^{2})(X_{1} − X_{2})^{−1}X_{1}. \end{aligned}  (3)

For n ≥ 3 it is impossible to write a formula valid for all independent matrices; for example, if n = 3, then

\begin{aligned}A_{1} = −((X_{1}^{3} − X_{2}^{3})(X_{1} & − X_{2})^{−1} − (X_{1}^{3} − X_{3}^{3})(X_{1} − X_{3})^{−1}) \\ & ((X_{1}^{2} − X_{2}^{2})(X_{1} − X_{2})^{−1} − (X_{1}^{2} − X_{3}^{2})(X_{1} − X_{3})^{−1})^{−1} \end{aligned}  (4)

provided that the right hand side exists (which does not follow from X_{1}, X_{2}, X_{3} being independent); otherwise the expression will be different.

An expression of A_{1}, …, A_{n} as functions of X_{1}, …, X_{n}, valid for generic independent X_{1}, …, X_{n}, may be given in terms of Gelfand–Retakh’s quasideterminants, which are defined as follows.

Definition 2.2 [GR]. Let A = {∥a_{ij}∥, i ∈ I, j ∈ J} be a square matrix of order n = card I = card J with formal non-commutative entries a_{ij}. For p ∈ I, q ∈ J denote by A^{pq} the submatrix {∥a_{ij}∥, i ∈ I − p, j ∈ J − q} of A. The formula

|A|_{pq} = a_{pq} − ∑_{{i∈I−p \atop j∈J−q}} a_{pj}|A^{pq}|_{ij}^{−1}a_{ip}

(which reduces to |A|_{pq} = a_{pq} if n = 1) defines inductively n^{2} quasideterminants |A|_{pq} of the matrix A. (In the commutative case |A|_{pq} = ±det A/det A_{pq}.)

Quasideterminants possess some basic properties of determinants; in particular, the following Kramer rule holds.

Proposition 2.3 [GR]. If (x_{1}, …, x_{n}) is the solution of a system of equations

∑_{j=1}^{n} a_{ij}x_{j} = ξ_{i} (i = 1, …, n),

then for any i

x_{j} = |A|_{ij}^{−1}|A_{j}(ξ)|_{ij},

where A = ∥a_{ij}∥ and A_{j}(ξ) is obtained from A be replacing its j-th column by the column (ξ_{1}, …, ξ_{n}).

Corollary 2.4. For generic independent X_{1}, …, X_{n} and for arbitrary i

−A_{j} =

| \begin{matrix}I & X_{1} & … & X_{1}^{n−1} \\ I & X_{2} & … & X_{2}^{n−1} \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ I & X_{n} & … & X_{n}^{n−1} \end{matrix} |_{i,n−j}^{−1} | \begin{matrix}I & X_{1} & … & X_{1}^{n−j−1} & X_{1}^{n} & X_{1}^{n−j+1} & … & X_{1}^{n−1} \\ I & X_{2} & … & X_{2}^{n−j−1} & X_{2}^{n} & X_{2}^{n−j+1} & … & X_{2}^{n−1} \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ I & X_{n} & … & X_{n}^{n−j−1} & X_{n}^{n} & X_{n}^{n−j+1} & … & X_{n}^{n−1} \end{matrix} |_{i,n−j} (5)

This expression times (−1)^{j} (with a minor change of notations and with i = n) is called in [GKLLRT], Section 7.1, the j-th elementary symmetric function in X_{1}, …, X_{n}. It is proved in [GKLLRT] and is obvious from Corollary 2.4, that for generic X_{1}, …, X_{n} it is really symmetric in X_{1}, …, X_{n}.

3. The case n = 2

Since for any square matrices X, Y

| \begin{matrix}I & I \\ X & Y \end{matrix} | = det (Y − X),

the following is precisely Theorem 1.1 for the case n = 2.

Proposition 3.1. Let A, B, X, Y be square matrices of the same order with X − Y being non-degenerate. If

\begin{aligned}X^{2} + AX + B & = 0, \\ Y^{2} + AY + B & = 0. \end{aligned}  (6)

Then

tr A = −(tr X + tr Y)
det B = det Xdet Y.

Proof. 1. The equalities (6) imply

−A = (X^{2} − Y^{2})(X − Y)^{−1},  (7)

and since X^{2} − Y^{2} = (X + Y)(X − Y) + XY − Y X, then

−A = X + Y + (XY − Y X)(X − Y)^{−1}.

Therefore

−tr A = tr X + tr Y + tr XY (X − Y)^{−1} − tr Y X(X − Y)^{−1},  (8)

and the relations

\begin{aligned}X(X − Y)^{−1} & = Y (X − Y)^{−1} + I, \\ (X − Y)^{−1}X & = (X − Y)^{−1}Y + I \end{aligned}

together with the identity tr UV = tr V U imply

tr XY (X − Y)^{−1} = tr Y (X − Y)^{−1}X = tr Y (X − Y)^{−1}Y + tr Y,
tr Y X(X − Y)^{−1} = tr X(X − Y)^{−1}Y = tr Y (X − Y)^{−1}Y + tr Y.

Hence tr XY (X − Y)^{−1} = tr Y X(X − Y)^{−1}, and (8) yields

−tr A = tr X + tr Y.

2. The first of the equalities (6) implies

B = −(X + A)X,  (9)

and using (7) we get

(X + A)(X − Y) = (X − (X^{2} − Y^{2})(X − Y)^{−1})(X − Y)
= X^{2} − XY − X^{2} + Y^{2}
= Y^{2} − XY = (X − Y)(−Y).

Therefore

X + A = (X − Y)(−Y)(X − Y)^{−1},
det (X + A) = det (−Y),

and in virtue of (9)

det B = det (−(X + A))det X = det Y det X.

4. The general case Unlike the above proof for n = 2, our proof in the general case is not reduced to a

direct calculation and uses the specifics of the matrix algebra.

Lemma 4.1. Let A_{1}, …, A_{n}, X_{1}, …, X_{n} be square matrices of some order k, and let the eigenvalues of the matrices X_{1}, …, X_{n} be kn pairwise different complex numbers. If

X_{i}^{n} + A_{1}X_{i}^{n−1} + … + A_{n} = 0  (10)

for i = 1, …, n, then

tr A_{1} = −(tr X_{1} + … + tr X_{n}),
det A_{n} = (−1)^{kn}det X_{1} … det X_{n}.

Proof. Let λ_{ij}, j = 1, …, k be eigenvalues of X_{i} and let X_{i}v_{ij} = λ_{ij}v_{ij}, v_{ij} ≠ 0. Then for arbitrary i, j (10) implies

\begin{aligned}0 & = (X_{i}^{n} + A_{1}X_{i}^{n−1} + … + A_{n})v_{ij} \\ & = (λ_{ij}^{n} + λ_{ij}^{n−1}A_{1} + … + A_{n})v_{ij}, \end{aligned}

whence

det (λ_{ij}^{n} + λ_{ij}^{n−1}A_{1} + … + A_{n}) = 0.

Obviously,

P (λ) = det (λ^{n} + λ^{n−1}A_{1} + … + A_{n})

is a monic polynomial in λ of degree kn, and since P (λ_{ij}) = 0 and all λ_{ij} are different, then

det (λ^{n} + λ^{n−1}A_{1} + … + A_{n}) = ∏_{i,j}(λ − λ_{ij}).  (11)

Equate constant terms and the coefficients in the term with λ^{kn−1} of the two sides of (11); we have

det A_{n} = ∏_{i,j}(−λ_{ij}) = (−1)^{kn} ∏_{i,j} λ_{ij} = (−1)^{kn} ∏_{i} det X_{i}.
trA_{1} = ∑_{i,j}(−λ_{ij}) = − ∑_{i} tr A_{i}.

Remark 4.2. Equating other coefficients of polynomials in (11), we may get kn − 2 more identities. Most of them involve matrices, obtained by combination of columns of different matrices A_{i}, but in the two extreme cases (those of λ^{p} with p = kn − 2 and 1) we get the formulas which are worth mentioning:

tr A_{2} + σ_{2}(A_{1}) = ∑_{1≤i<j≤n}(tr X_{i}tr X_{j}) + ∑_{1≤i≤n} σ_{2}(X_{i});  (12)

det A_{n−1} · tr A_{n}A_{n−1}^{−1} = (−1)^{kn−1}(∏ det X_{i})(∑ tr X_{i}^{−1}),  (13)

where σ_{2} in (12) denotes the second coefficient (the coefficient in the term with λ^{k−2}) of the characteristic polynomial of a matrix, and (13) is valid only if A_{n−1}, X_{1}, …, X_{n} are non-degenerate.

Proof of Theorem 1.1. Let U, V ⊂ Mat_{k}ℂ×…×Mat_{k}ℂ (n factors) are respectively the set of all n-tuples (X_{1}, …, X_{n}) of independent matrices and the set of all n-tuples (X_{1}, …, X_{n}) of matrices whose eigenvalues are kn pairwise different complex numbers. Obviously the both sets are Zariski open and non-empty, hence U ∩ V is dense in U. According to Lemma, the equalities (2) both hold in U ∩ V. Since the both sides of each of the equalities (2) are continuous on U (with respect to X_{1}, …, X_{n}), the equalities (2) hold on the whole U.

5. A further generalization

Let R be an associative ring with unity, and k be a field. Suppose that there fixed either an additive homomorphism tr: R → k satisfying the condition trUV = trV U for any U, V ∈ R, or a ring homomorphism det: R → k (or both).

Proposition 5.1. Let A, B, X, Y ∈ R with X −Y being invertible. If X^{2} +AX +B = 0, Y^{2} + AY + B = 0, then tr A = −(tr X + tr Y) and/or det B = det Xdet Y, whichever is defined.

The proof is the same as in Section 3. To generalize to arbitrary rings the general case of Theorem 1.1 we need an explicit

expression of A_{1}, A_{n} via X_{1}, …, X_{n}.

Theorem 5.2. Let

\begin{aligned}A_{1} & = a_{1}(X_{1}, …, X_{n}), \\ A_{n} & = a_{n}(X_{1}, …, X_{n}) \end{aligned}

be expressions of A_{1}, A_{n} via X_{1}, …, X_{n} involving ring operations and taking inverses and valid where these inverses exist (like (3), (4), (5)). Then for any X_{1}, …, X_{n} ∈ R each of the equalities

tr a_{1}(X_{1}, …, X_{n}) = −(tr X_{1} + … + tr X_{n}),
det a_{n}(X_{1} …, X_{n}) = det (−X_{1}) … det (−X_{n})

holds provided that the both sides exist (that is tr or det is defined and the inverses involved in a_{1} or a_{n} exist in R).

Theorem 5.2 cannot be proved by arguments similar to that of Section 4. But it is known that identities which hold for complex matrices hold also in arbitrary associative rings with unities (see [A], Section 12.4.3). Hence Theorem 5.2 follows from Theorem 1.1.

Acknowledgement. We are indebted to C. Itzykson and I. Kaplansky for interesting discussion.

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