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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.

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  • $\begingroup$ Can anyone tell me why my question is closed or how I can modify the content so that it can be reopened? Thanks $\endgroup$
    – Ivan Lg
    Nov 6, 2020 at 10:14
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    $\begingroup$ adding some commentary on where this problem came from and what background knowledge you have would be helpful. For example, you said "If 𝑋 is a normal matrix, show that the inequality is still true. I have no idea how to start with the second problem." but to most people the second part is an immediate corollary of the first since normal matrices are unitarily diagonalizable and the Frobenius norm is unitarily invariant. $\endgroup$ Nov 8, 2020 at 19:59
  • $\begingroup$ Thanks for your reply. At the beginning, I keep trying to use the definition of normal matrix $AA^*=A^*A$, and thinking how I can use this definition in the calculation of norm, and that's why I don't know how to solve the problem. Once you tell me about the property that normal matrices are unitary diagonalizable, then I know that I make a silly mistake and the second problem actually is very simple. Thanks very much. $\endgroup$
    – Ivan Lg
    Nov 16, 2020 at 13:13
  • $\begingroup$ Wow! this fact (without the $X$ being diagonal assumption) was only proven in 2008 tu-chemnitz.de/mathematik/preprint/2008/PREPRINT_05.pdf $\endgroup$
    – Condo
    Jan 11, 2021 at 22:54

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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.

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