Prove $\min_{||X||_2=1}||BX-XC||_2\le\min\{|\mu-\lambda|;\mu\in\lambda(B),\lambda\in\lambda(C)\}$ as $B,C\in M_{n\times n}$ have different eigenvalues For any matrices $B\in\mathbb C^{n\times n},C\in\mathbb C^{m\times m}$, define the separation of matrices $$sep(B,C):=\min_{X\in\mathbb C^{n\times m},||X||_2=1}||BX-XC||_2.$$
Prove that if $\lambda(B)\cap\lambda(C)=\emptyset$, then we have $$sep(B,C)\le\min\{|\mu-\lambda|;\mu\in\lambda(B),\lambda\in\lambda(C)\},$$ and  the equality holds for normal $B,C$.
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My first idea is that when $B,C$ are normal, we may assume $B=U\Lambda_BU^H,C=V\Lambda_CV^H$ where $U,V$ are unitary and we have $sep(B,C)=\min_{X\in\mathbb C^{n\times m},||X||_2=1}||\Lambda_BX-X\Lambda_C||_2$ by properties of 2-norm. Since $\Lambda_BX-X\Lambda_C=((b_i-c_j)x_{ij})_{i,j=1}^n$, when $|b_i-c_j|$ becomes minimum we can choose $x_{pq}=\begin{cases}1,~\text{ if }(p,q)=(i,j)\\0.5,~\text{ if }p-q\equiv i-j\pmod n\\0,~\text{others}\end{cases}$ to achieve the equality. However this doesn't solve any part of the problem, and I cannot come up with more ideas. Are there any possible ways of proceeding further? Thanks!
Edit1. By properties of Schur decomposition we may assume that (in general case) $B=UT_BU^H,C=VT_CV^H$ where $U,V$ are unitary and $T_B,T_C$ being upper triangle, we may also assume that $|{T_B}_{11}-{T_C}_{11}|=argmin_{\mu,\lambda}\{|\mu-\lambda|;\mu\in\lambda(B),\lambda\in\lambda(C)\}$, and $sep(B,C)=\min_{X\in\mathbb C^{n\times m},||X||_2=1}||T_BX-XT_C||_2$ Then if we let $x_{pq}=\begin{cases}1,~\text{ if }(p,q)=(1,1)\\0,~\text{others}\end{cases}$ we have $||X||_2=1$ and $||T_BX-XT_C||_2=\min\{|\mu-\lambda|;\mu\in\lambda(B),\lambda\in\lambda(C)\}$. Hence the inequality is proved. But how to prove the equality holding case?
 A: Let $u$ be a unit right-eigenvector corresponding to an eigenvalue $\mu_j$ of $B$ and $v$ be a unit left-eigenvector corresponding to an eigenvalue $\lambda_i$ of $C$. Then
$$
\min_{\|X\|_2=1}\|BX-XC\|_2
\le\|Buv^\top-uv^\top C\|_2
=|\mu_j-\lambda_i|\|uv^\top\|_2
=|\mu_j-\lambda_i|.
$$
Hence $\min_{\|X\|_2=1}\|BX-XC\|_2\le\min_{i,j}|\mu_j-\lambda_i|$. Note that this is true regardless of any overlapping (or the lack of it) between the spectra of $B$ and $C$.
The claim about the normal case is false. Here is a random counterexample generated by computer:
\begin{aligned}
B&=\pmatrix{0.9926&0\\ 0&0.4846},\\
C&=\pmatrix{0.1725&0\\ 0&0.7319},\\
X_0&=\pmatrix{0.092101&-0.682802\\ -0.647583&0.511843},\,\sigma_1(X_0)=1,\,\sigma_2(X_0)=0.3950,\\
BX_0-X_0C&=\pmatrix{0.075535&-0.178049\\ -0.202120&-0.126554},\\
\|BX_0-X_0C\|_2&=0.2435,\\
\min_{i,j}|\mu_i(B)-\lambda_j(C)|
&=\min\{0.8201,\,0.2607,\,0.3121,\,0.2473\}=0.2473.
\end{aligned}
Strict inequality holds in this case:
$$
\min_{\|X\|_2=1}\|BX-XC\|_2
\le\|BX_0-X_0C\|_2
=0.2435<0.2473
=\min_{i,j}|\mu_i(B)-\lambda_j(C)|.
$$
