What is the infimal norm of a matrix in a conjugacy class? $\newcommand{\tr}{\operatorname{tr}}$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 $\mathrm{GL}(d, \mathbb{C})$.
 A: @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, $tr(B^TB)=tr(T^*T)> \sum_j |T_{j,j}|^2=\sum_j |\lambda_j|^2$.  $\square$
A: 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.
