Suppose $A=(a_{ij})$ is an $n\times n$ complex matrix, $\operatorname{tr}(A)=\displaystyle \frac{1}{n}\sum_{i}a_{ii}$. I wonder whether there exists a relationship between $\|A\|$ (the operator norm) and $\operatorname{tr}(A^k)$, where $k=1,\cdots,n$?

  • $\begingroup$ in general the trace is not divided by $n$, this is a especial trace operator or it is a typo? $\endgroup$ – Masacroso Nov 8 '18 at 19:09
  • $\begingroup$ @Masacroso: normalizing the trace so that it becomes a state is a very standard thing. $\endgroup$ – Martin Argerami Nov 9 '18 at 0:17

In general no, as nilpotent operators exist. For instance consider $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}. $$ Then $\|A\|=1$ and $\operatorname{tr}(A^k)=0$ for all $k$.

When $A$ is normal, it is easy to check that $$\tag1 \|A\|=\lim_{k\to\infty}|n\operatorname{tr}(A^k)|^{1/k}. $$ Indeed, you have that $\|A\|=\max\{|\lambda_j|:\ j=1,\ldots,n\}$, where $\{\lambda_j\}$ are the eigenvalues of $A$ counting multiplicities. If we take $\lambda_1$ to be the one with $|\lambda_1|=\|A\|$,
\begin{align} |\operatorname{tr}(A^k)|^{1/k}&=\exp\left({\tfrac1k\,\log|\sum_{j=1}^n\lambda_j^k|}\right) =\exp\left({\tfrac1k\,\log|\lambda_1^k|\,|1+\sum_{j=1}^n(\lambda_j/\lambda_1)^k|}\right)\\ \ \\ &=|\lambda_1|\,\exp\left({\tfrac1k\,\log |1+\sum_{j=1}^n(\lambda_j/\lambda_1)^k|}\right)\xrightarrow[k\to\infty]{}|\lambda_1|,\\ \ \\ \end{align} as the expression inside brackets is bounded above by $\tfrac1k\,\log(1+n)$.

  • $\begingroup$ why the second equality has no the term $1/n$? $\endgroup$ – mathrookie Nov 9 '18 at 15:09
  • $\begingroup$ I forgot an $n$ in $(1)$. $\endgroup$ – Martin Argerami Nov 9 '18 at 15:13
  • $\begingroup$ So in your proof,$tr$ denotes the sum of diagonal elements of a matrix? $\endgroup$ – mathrookie Nov 9 '18 at 15:20
  • $\begingroup$ No. $ \ \ \ \ $ $\endgroup$ – Martin Argerami Nov 9 '18 at 15:21
  • $\begingroup$ $|tr(A^k)|=1/n|\sum \lamda j k$? $\endgroup$ – mathrookie Nov 9 '18 at 15:26

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