# operator norm of a matrix and trace of a matrix

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 $$k=1,\cdots,n$$?

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

## 1 Answer

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 $$\tfrac1k\,\log(1+n)$$.

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