Show that $\|XY-YX\|^2\le2\|X\|^2\|Y\|^2$ for diagonal matrix $X$ and any matrix $Y\in M_n$

Question

I am studying on the topic of norm of the commutator and I would like to solve the following problems:

For any matrix $X\in M_n$, $\|X\|=\sqrt{\sum|x_{ij}|^2}=\sqrt{trA^*A}$ , where $A^*$ is the conjugate transpose of $A$(So the field is $\mathbb C$).

Let $X=diag(x_1,x_2,...,x_n)$ be a diagonal matrix in $M_n$. Show that $\|XY-YX\|^2\le2\|X\|^2\|Y\|^2$ for any matrix $Y\in M_n$.

This is the first part of my problem. I try the question in this way:

$X=\operatorname{diag}(x_1,x_2,...,x_n)$, $Y=\begin{bmatrix}y_{11} & y_{12} & ... & y_{1n}\\y_{21} & y_{22} & ... & y_{2n}\\ \vdots & \vdots & \ddots & \vdots \\y_{n1} & y_{n2} & ... & y_{nn}\end{bmatrix}$

then $$XY-YX=\begin{bmatrix}(x_1-x_1)y_{11} & (x_1-x_2)y_{12} & ... & (x_1-x_n)y_{1n}\\(x_2-x_1)y_{21} & (x_2-x_2)y_{22} & ... & (x_2-x_n)y_{2n}\\ \vdots & \vdots & \ddots & \vdots \\(x_n-x_1)y_{n1} & (x_n-x_2)y_{n2} & ... & (x_n-x_n)y_{nn}\end{bmatrix}$$

So $\|XY-YX\|^2=\sum|(x_i-x_j)y_{ij}|^2$ for $1\le i,j\le n$ and I get stuck in this step.

The second part of my problem is:

If $X$ is a normal matrix, show that the inequality is still true. I have no idea how to start with the second problem.

Answer 1

The slightly more general version of your question was in fact Conjectured back in 2005 for arbitrary matrices (without the assumption that $X$ is diagonal) by Bottcher and Wenzel. It was proven shortly after by Vong and Jin, and you can see the follow-up paper by the authors of the conjecture here with the proof of the desired fact.

The proof by Bottcher and Wenzel doesn't appear to be simple enough to explain in an MSE post. That being said I will show you how elementary methods can get you a weaker bound that is off by a factor of only $2$.

The matrix norm you mention is called the Frobenius norm, so we will denote it by $\|\cdot\|_F$. To obtain a slightly weaker bound, you can use the fact that $(M_n(\mathbb{C}),\|\cdot\|_F)$ is a Hilbert space (with the Frobenius a.k.a Hilbert-Schmidt inner-product), and therefore the parallelogram law holds. In other words, for any complex $n\times n$ matrices $X,Y$ we have that $$\|XY-YX\|_F^2+\|XY+YX\|_F^2=2\|XY\|_F^2+2\|YX\|_F^2.$$

Another useful fact is that the Frobenius norm is submultiplicative and therefore $$\|XY\|_F\leq \|X\|_F\|Y\|_F,$$ for any $X,Y\in M_n(\mathbb{C})$.

Combining these two facts we can see that for any matrices $X$ and $Y$ $$\|XY-YX\|_F^2\leq 2\|X\|_F^2\|Y\|_F^2+2\|Y\|_F^2\|X\|_F^2=4\|X\|_F^2\|Y\|_F^2.$$

The above bound is weaker than the bound in the question (which is in fact tight), however, it can be proven using elementary results.

Remark 1. I have not used the fact that $X$ is diagonal... perhaps the argument in the proof is easier with this assumption, however, this was not obvious...

Remark 2 As mentioned in the comments the Frobenius norm is unitarily invariant, and therefore if $U\Sigma V$ is the singular value decomposition of $X$ with $\Sigma$ being the diagonal matrix of singular values then $\|X\|_F=\|U\Sigma V\|_F=\|\Sigma\|_F$, in particular, if $X$ was normal then $V=U^*$, and the same thing would essentially hold.