An upper-bound for $\text{trace}(A^2)$ in terms of $\text{trace}(A)$

Question

I am looking for an upper-bound for $\text{trace}(A^2)$ in terms of $\text{trace}(A)$. Matrix $A$ is Hermititan, but not positive-definite, so unfortunately I cannot use $\text{trace}(A^2)\leq \text{trace}(A)^2$.

Answer 1

The trace of a matrix is equal to the sum of its eigenvalues, so

$$\text{trace}(A)=\sum_{i=1}^n\lambda_i \qquad \text{trace}(A^2)=\sum_{i=1}^n\lambda_i^2$$

Where the eigenvalues $\lambda_i$ are real since $A$ is assumed to be hermitian. In particular, you won't be able to bound $\text{trace}(A^2)$ by $\text{trace}(A)$ since for example $A_w =\begin{pmatrix}w&0\\0&-w\end{pmatrix}$ has $\text{trace}(A_w)=0$, but $\text{trace}(A_w^2)=2w^2$ for all $w$.