How do I show that $\inf\limits_{\det(X)\neq0}\|X^{-1}AX\|^{2}_{F}=\sum\limits_{\lambda\in{\Lambda}}|\lambda|^{2}$? Show that
$$\inf\limits_{\det(X)\neq 0}\|X^{-1}AX\|^2_F=\sum_{\lambda\in\Lambda}|\lambda|^{2}$$
holds, where $\Lambda(A)$ is the set containing all eigenvalues of A, and $\|\cdot\|_{F}$ is the Frobenius norm.
Also, I want to show that it is indeed the minimum if and only if $A$ is diagonalizable. 
 A: First, there is a useful relation between eigenvalues and the Frobenius norm$\color{red}{^*}$:
$$
\sum_i|\lambda_i|^2\leq\|A\|_F^2,
$$
where $\lambda_i$ ($i=1,\ldots,n$) are the eigenvalues of $A$. Since for any nonsingular $X$, $X^{-1}AX$ has the same eigenvalues as $A$, we have
$$
\sum_i|\lambda_i|^2\leq\|X^{-1}AX\|_F^2.
$$
This gives us:

For any nonsingular $X$ and $A$ with eigenvalues $\lambda_1,\ldots,\lambda_n$, 
  $$\tag{1}
\left(\sum_{i}|\lambda_i|^2\right)^{1/2}\leq\|X^{-1}AX\|_F.
$$

Next, if $A$ is diagonalizable, it is easy to see that with $A=Y\Lambda Y^{-1}$, the equality in (1) is attained with $X=Y^{-1}$.
If $A$ is not diagonalizable, a proper scaling of the Jordan form helps here. We know that there is a nonsingular $Y$ and a Jordan matrix $J$ such that $A=YJY^{-1}$, where the Jordan blocks in $J$ have the form
$$
J_{\lambda}=\lambda I + N,
$$
where $N$ is zero elsewhere except ones on the first superdiagonal. Now take any $\epsilon>0$ and consider $D_{\epsilon}:=\mathrm{diag}(1,\epsilon,\ldots,\epsilon^{n-1})$ and observe that
$$
D_{\epsilon}^{-1}J_{\lambda}D_{\epsilon}=\lambda I + \epsilon N.
$$
This indicates that, by a proper diagonal similarity transformation, you can make the nilpotent part of $J_{\lambda}$ arbitrarily small and this applies also to the whole Jordan matrix. Summarizing:

For any matrix $A$ and $\epsilon>0$, there is a nonsingular $Y$ such that
  $$
A=YJY^{-1}, 
$$
  where $J=\Lambda+N$, $\Lambda$ is diagonal (containing the eigenvalues of $A$ on the diagonal), and $N$ is such that $\|N\|_F\leq\epsilon$.

So, setting again $X=Y^{-1}$ in (1), you can make the right-hand side of the inequality arbitrarily close to the left-hand side. Nevertheless, the infimum is not attained if $A$ is not diagonalizable.

$\color{red}{^*}$ The fact that 
$$\sum_i|\lambda_i|^2\leq\|A\|_F^2$$
can be shown, e.g., using the Schur form. Let $A=QTQ^*$ be the Schur decomposition of $A$ where $Q$ is unitary and $T$ upper triangular with the eigenvalues $\lambda_i$ on the diagonal. Since the eigenvalues can come in any prescribed order on the diagonal of $\lambda_i$, say, the row/column $i$ of $T$ contains $\lambda_i$ on the diagonal. If $t_i$ is the $i$th column of $T$, we have hence $|\lambda_i|\leq\|t_i\|_2$. Therefore, $$\sum_i|\lambda_i|^2\leq\sum_i\|t_i\|_2^2=\mathrm{trace}(T^*T)=\|T\|_F^2=\|A\|_F^2.
$$
