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I've been working out a question in Applied Numerical Linear Algebra by James W. Demmel in Chapter 4 but totally stuck. The question is: let $A$ be any n-by-n matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$. Prove that

$$ \sum_{i=1}^n |\lambda_i|^2 = \min_{\det(S)\neq 0} \|SAS^{-1}\|_F^2$$

I've been trying to use the formula $\|A\|_F^2 = \text{trace}(AA^*)$, but this doesn't seem to help.

Here's what I did:

$$\|S^{-1}AS\|_F^2 = \text{trace}(S^{-1}AS(S^{-1}AS)^*) = \text{trace}(S^{-1}ASS^*A^*S^{-*})=\text{trace}(ASS^*A^*(SS^*)^{-1})$$

(where $A^{-*}$ means inverse of conjugate transpose, a convention highly abused in the book)

But how do we handle the $SS^*$ and its inverse? Please, I need some help.

Henry Davii
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1 Answers1

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I have a proof where we replace $\min$ by $\inf$:

$$ \sum_{i=1}^n |\lambda_i|^2 = \inf_{\det S \ne 0} \| SAS^{-1} \|_{\rm F}^2. $$

I would be interested to see a proof that is able to replace $\min$ by $\inf$.

We prove this equality by proving two inequalities. To show "$\ge$", note that the standard construction of the Jordan normal form allows one to have $\delta$'s on the off-diagonal in place of $1$, where $\delta \ne 0$ is any nonzero real number. Thus, by choosing $S = S_\delta$ to produce a "$\delta$-Jordan normal form" for $A$, $S_\delta A S_\delta^{-1}$ is a bidiagonal matrix with the eigenvalues of $A$ on the diagonal and only $0$ and an arbitrarily small number $\delta$ on the off-diagonal. Thus, $\|S_\delta A S_\delta^{-1}\|_{\rm F}^2 \le \sum_{i=1}^n |\lambda_i|^2 + (n-1)\delta^2$. Taking $\delta \to 0$ establishes that $\sum_{i=1}^n |\lambda_i|^2 \ge \inf_{\det S \ne 0} \| SAS^{-1} \|_{\rm F}^2$. Note that the $\inf$ is a $\min$ if $A$ is diagonalizable.

To prove the "$\le$" part, we use the fact that $\sum_{i=1}^n |\lambda_i(B)|^2 \le \operatorname{tr} (B^*B) = \|B\|_{\rm F}^2$ for any matrix $B$; see this question for a proof. Instantiate this result with $B = SAS^{-1}$ and note that $A$ and $SAS^{-1}$ have the same eigenvalues. Thus,

$$ \sum_{i=1}^n |\lambda_i|^2 \le \| SAS^{-1} \|_{\rm F}^2 \mbox{ for any } S \mbox{ nonsingular}. $$

This proves "$\le$" and thus completes our proof of equality.

eepperly16
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  • Firstly, thank you for the excellent answer. I'm wondering is there a possibility that $\min$ is not attainable if $A$ is not normal? I'm speculating this because$GL(n,\mathbb{C})$ is not closed, there need not be a minimum. – Henry Davii Dec 02 '21 at 06:27
  • @HenryDavii Picking $S$ to diagonalize $A$, the $\min$ is attained whenever $A$ is diagonalizable. I believe is that this is an if, and only if—that is, if $A$ is not diagonalizable, then the $\min$ is not attained. For proof, note that the inequality $\sum_{i=1}^n |\lambda_i(B)|^2 \le |B|{\rm F}^2$ has equality if, and only if, $B$ is normal (see the linked MSE question). But a non-diagonalizable matrix is not similar to a normal matrix, so $\sum{i=1}^n |\lambda_i|^2 < |SAS^{-1}|_{\rm F}^2$. – eepperly16 Dec 02 '21 at 18:14