in going through some books about numerical mathematics I found the following exercise:

Let $A,B \in \mathbb{R}^{n\times n}$ with $A$ symmetrical and rank($A$) = rank(B) = $n$. Define $M = \left[\begin{matrix} A & B \\ B^T & 0 \end{matrix}\right] $.

The statement now is, that $M$ has exactly $n$ positive and $n$ negative eigenvalues. And to prove it one should use the Courant minimax principle.

First we can see, that $M$ ist symmetrical, so Courant minimax principle is actually applicable for $M$. As the eigenvalues given by the minimax principle are ordered ($\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_{2n}$) we only have to consider two of them: $\lambda_n$ and $\lambda_{n+1}$. If $\lambda_n$ is positive and $\lambda_{n+1}$ negative, then the statement is proven right.

And this is where my problems begin. The Courant minimax principle gives that $\lambda_k = \max\limits_{U \in \mathcal{U}_k} \min\limits_{x \in U, ||x|| = 1}{x^T M x}$, where $\mathcal{U}_k = \{ U \subset \mathbb{R}^{2n} | \dim{U} = 2n + 1 -k\}$. I can't figure out how to make any kind of estimate for the value of $\lambda_k$ (in particular for $k = n$ and $k = n+1$) given how little we know about the matrix $M$ or it's components $A$, and $B$.


2 Answers 2


Note. The OP has gotten the principle wrongly. What he/she states actually holds when the eigenvalues are arranged in ascending order, not in descending order. In other words, if we denote the eigenvalues of a $2n\times2n$ Hermitian matrix $M$ by $\lambda_1(M)\le\cdots\le\lambda_{2n}(M)$, then $$ \lambda_k(M) = \max_{\dim U=2n+1-k}\ \min_{x\in U,\ \|x\|=1} x^\ast Mx. $$

Now, return to the problem, where $M=\pmatrix{A&B\\ B^T&0}$ with $A$ is real symmetric and $B$ is real and nonsingular. Clearly $M$ is nonsingular too. So, it suffices to prove that $\lambda_{n+1}(M)\ge0$: as $M$ has no zero eigenvalues, if $\lambda_{n+1}(M)\ge0$, we must have $\lambda_{n+1}(M)>0$. Yet, $-M$ has the same structure as $M$. So, by the same reasoning, we also have $\lambda_{n+1}(-M)>0$, i.e. $\lambda_{n}(M)<0$.

To show that $\lambda_{n+1}(M)\ge0$, Courant-Fischer principle tells us to prove that $\max_{\dim U=n}\ \min_{x\in U,\ \|x\|=1} x^\ast Mx\ge0$, but this is true because $x^\ast Mx$ is identiacally zero on the $n$-dimensional vector subspace $$ U_0=\left\{x=\pmatrix{0\\ v}:v\in\mathbb C^n\right\}. $$


(Too long for a comment.) I don't know why we "should" apply the Courant-Fischer minimax principle. The statement can be proved pretty easily by Sylvester's law of inertia (which is a much weaker statement than the Courant-Fischer minimax principle).

Since $B$ has full rank, by matrix congruence (consider $\pmatrix{I&-\frac12A(B^T)^{-1}\\ 0&I}$), you may assume that $A=0$. In this case, the block matrix is clearly nonsingular. Now, if $(x^T,y^T)$ (with $x,y\in\mathbb R^n$) is an eigenvector for a nonzero eigenvalue $\lambda$, then $(x^T,-y^T)$ is an eigenvector for the eigenvalue $-\lambda$. So, eigenvalues of the block matrix occur in pairs of nonzero values with equal moduli but opposite signs. Hence the result.

  • $\begingroup$ Sorry for the late reply! Your answer nicely shows my original statement, though I wouldn't say we can assume $A=0$, because $A$ has full rank by definition. Sadly this does not help me in bettering my understanding of the Courant minimax principle. $\endgroup$
    – Benedikt
    Jan 27, 2015 at 14:05

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