The sum of square of eigenvalues equals minimal Frobenius norm under similar transformation

Solution 1:

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.