How to show $\sqrt{\text{Tr}(A^2)} \leq \text{Tr}(A)$?

Question

Let $A$ be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally, $$\sqrt{\text{Tr}(A^2)} \leq \text{Tr}(A)$$ Also, show when $A$ is an $n \times m$ the following is true $$\sqrt{\text{Tr}(A^TA)} \leq \|A\|_*$$ where $\|\cdot\|_*$ is nuclear norm which is the summation of the singular values.

Answer 1

For the first you can assume that $A$ is diagonal with diagonal entries $a_1,\ldots,a_n$, all $\ge0$. Then your inequality becomes $$\sum_{i=1}^n a_i^2\le\left(\sum_{i=1}^n a_i\right)^2$$ which is clearly true on expanding the right side, recalling all variables are non-negative.

Answer 2

Let $\sigma_1, \ldots, \sigma_r$ be the singular values of $A$. Then $$\sqrt{\text{Tr}(A^\top A)} = \sqrt{\sum_i \sigma^2_r} \le \sum_i |\sigma_r| = \|A\|_*.$$