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$\newcommand{\tr}{\operatorname{tr}}$Let $A$ be a real $d \times d$ matrix. What is $$ F(A):=\inf_{M \in \text{GL}(d, \mathbb{R})} |MAM^{-1}|^2?$$ Here $| \cdot |$ is the standard Euclidean (Frobenius) norm. Is the infimum always obtained? I would like to find a simple formula for $F(A)$ in terms of $A$.

Cauchy-Schwarts inequality implies that $$( \tr(A))^2 \le d |A|^2$$ with equality if and only if $A$ is a multiple of the identity matrix.

Since the trace is an invariant of a conjugacy class, we deduce

$$ F(A) \ge \frac{\tr(A)^2}{d},$$ and this value is obtained if and only if $A$ is a multiple of the identity matrix.

Comment: Note that I am referring to conjugacy classes in $\mathrm{GL}(d, \mathbb{R})$, not in $\mathrm{GL}(d, \mathbb{C})$.

Asaf Shachar
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  • This vaguely reminds me of this post. However, that maximization problem is on $O(d)$, while you consider $GL(d)$, which is not compact. This can be a significant difficulty. – Giuseppe Negro Feb 12 '18 at 11:18

2 Answers2

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@Dunham was faster than me. I agree with the essential. Yet, the inequality to prove is

$F(A)\leq \sum_j |\lambda_j|^2$ where $spectrum(A)=(\lambda_j)_j$. In particular, if $A$ is nilpotent, then $F(A)=0$, and consequently, $F$ is not a norm.

Proof. Case 1. $spectrum(A)\subset \mathbb{R}$. (as Dunham). For every $\epsilon >0$, there is a $M\in GL(\mathbb{R})$ s.t. $MAM^{-1}=D+T$ where $D=diag(\lambda_i)$ and $T$ is strictly upper triangular and $||T||<\epsilon$ where $||.||$ is a matricial norm invariant by transposition. Note that $trace(DD^T)=\sum_j \lambda_j^2$; after it's easy.

Case 2. Otherwise, we may assume, by Case 1, that $A$ has only non-real eigenvalues of the form $(a_j\pm ib_j)_j$; then (cf. the standard reference given by Asaf Shachar), for every $\epsilon >0$, there is $M\in GL(\mathbb{R})$ s.t.

$MAM^{-1}=diag(U_1,\cdots,U_k)+T=\Delta+T$ where $U_j=\begin{pmatrix}a_j&b_j\\-b_j&a_j\end{pmatrix}$ and $T$ is strictly upper triangular and $||T||<\epsilon$ where $||.||$ is a matricial norm invariant by transposition. Note that $trace(\Delta\Delta^T)=\sum_j (a_j^2+b_j^2)=\sum_j|\lambda_j|^2$. After, it's easy. $\square$

EDIT 1. Proposition. $F(A)= \sum_j |\lambda_j|^2$.

Proof. It remains to show that if $spectrum(B)=(\lambda_j)_j$, then $\sum_j |\lambda_j|^2\leq tr(B^TB)$. There is a unitary matrix $U$ s.t. $B=UTU^*$ where $T$ is complex upper triangular. Then $tr(B^TB)=tr(UT^*U^*UTU^*)=tr(T^*T)\geq \sum_j |T_{j,j}|^2=\sum_j |\lambda_j|^2$. $\square$

EDIT 2. Proposition. The lower bound is reached IFF $A$ is diagonalizable over $\mathbb{C}$.

Proof. If $A$ is diagonalizable over $\mathbb{C}$, then $A$ is similar over $\mathbb{R}$ to $D$ (Case 1) or $\Delta$ (Case 2) and we are done.

Otherwise, let $M\in GL_n(\mathbb{R})$ and $B=MAM^{-1}$ (cf. EDIT 1). Since $T$ cannot be diagonal, $tr(B^TB)=tr(T^*T)> \sum_j |T_{j,j}|^2=\sum_j |\lambda_j|^2$. $\square$

  • Thanks. There is something I do not understand though: $\text{trace}(D^TD)=\text{trace}(A^TA)=\sum_i \sigma_i^2$ is the sum of the squared singular values of $A$, right? yet you wrote the sum of squares of the eigenvalues. Are they always the same? – Asaf Shachar Feb 22 '18 at 16:29
  • No. Take $\epsilon =1/k$. Then, there is a sequence $(M_k)$ s.t. ${||M_kAM_k^{-1}||_F}^2$ tends to ${||D||_F}^2=\sum_j\lambda_j^2$. –  Feb 22 '18 at 16:39
  • Sorry, I was confused. – Asaf Shachar Feb 22 '18 at 16:40
  • Oh, one more question: Do you know when the minimum is obtained? i.e. for which matrices $A$, $\sum_i \sigma_i^2=|A|^2=\sum_i \lambda_i^2$? This holds for any symmetric matrix, and for any conformal matrix, but is that all? (are these the only cases of equality?). Thanks for your efforts. – Asaf Shachar Feb 23 '18 at 06:49
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Not a proof, but an idea:

Claim 1: It suffices to consider $A$ in Jordan canonical form (upper triangular)

Claim 2: $F(A)\leq \|A \|_2^2$

Proof: Consider the diagonal matrix $D_t=\mathrm{diag}(t,t^2,\dots,t^d)$ for $t>0$. We compute $D_t^{-1}AD_t$. The diagonal still has the eigenvalues, while the off-diagonal terms are multiplied by a positive power of $t$. Letting $t\rightarrow 0$ we have $D_t^{-1}AD_t$ converging to a diagonal matrix with the eigenvalues on the diagonal, and the result follows.

Note: $\| A\|_2^2=\frac{\mathrm{tr}(A)^2}{d}$ for multiples of the identity. My guess is that $F(A)= \|A \|_2^2$ and the infimum is not a min unless the Jordan form of $A$ is diagonal.

Note: I have assumed the inf is over $\mathrm{GL}(d, \mathbb{C})$. If instead you meant $\mathrm{GL}(d, \mathbb{R})$, then this is all probably useless.

Dunham
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  • Thanks, this is interesting. Yeah, I meant infimum over real matrices. However, maybe we can use real Jordan forms in some way .(https://en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices). – Asaf Shachar Feb 22 '18 at 06:31