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

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