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I'm trying to prove that the trace of a matrix $\tau:\mathcal M_{n\times n}(\mathbb R)\to \mathbb R$ is a bounded operator, i.e. there is $K>0$ s.t. $|\tau(M)|\leq K\|M\|,$ where $$\|M\|=\sup_{|x|_2= 1}|Mx|_2,$$ where $|\cdot |_2$ denote the euclidian norm of $\mathbb R^n$. Find $$\|\!|\tau\|\!|:=\sup_{\|M\|=1}|\tau(M)|.$$

I'm very confused with all these norms.

Attempts

  1. For the fact that $\tau$ is continuous, it's quite easy since $$|\tau(M)|\leq |M_{11}|+...+|M_{nn}|\leq n^2 \mathcal N_1(M),$$ where $$\mathcal N_1(M)=\sum_{1\leq i,j\leq n}|M_{ij}|.$$ I know that $\mathcal N_1$ is a norm on $\mathcal M_{n\times n}(\mathbb R)$, and since all norms on $\mathcal M_{n\times n}(\mathbb R)$ are equivalent, there is $P>0$ s.t. $\mathcal N_1(M)\leq P\|M\|$. Therefore, setting $K=n^2P$, we get $$|\tau(M)|\leq K\|M\|,$$ as wished.

  2. I have no idea on how computing $|\!\|\tau\|\!|$. I suspect that $|\!\|\tau\|\!|=n^2$, but I have no idea on how to compute it. Any idea ?

David C. Ullrich
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joshua
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1 Answers1

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I think C-S yields

$$ |\tau(M)| \le \sum_{i=1}^n|\langle Me_i,e_i\rangle| \le \sum_{i=1}^n |Me_i|_2 |e_i|_2 \le \sum_{i=1}^n\|M\| = n \|M\|. $$

For $M=I_n$ we have $\tau(M)=n$ and $\|M\| =1$. Hence $||| \tau |||=n$.

Gerd
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    Please search for duplicates before answering! (=: – PinkyWay Aug 23 '22 at 15:28
  • @SummerChild In fact I did. Searched under "trace norm" and found the corresponding answer with respect to the Frobenius norm, but not exactly the version above. I agree that its maybe a dublicate, but I don't search for hours. – Gerd Aug 23 '22 at 15:41
  • The search took me less than a minute, I just entered: "finite dimensional" AND "bounded" AND "operator". – PinkyWay Aug 23 '22 at 16:17
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    @SummerChild Yes, the link you found shows that $\tau$ is continuous, but say's nothing on the quantity of $|||\tau|||$. In the original question $|||\tau|||=n^2$ was presumed, which is wrong. This was the reason for my answer. – Gerd Aug 23 '22 at 18:11