Let $A$ be a symmetric matrix which has $k$ non-zero eigenvalue. Show that the square of Frobenius norm is always bigger than the average of squared eigenvalues. That is:
$$\|A\|_F^2 \geq \frac{1}{k} (\sum_{i=1}^k |\lambda_i|)^2$$
My try: applying $\|A\|_F^2=\sum_{i=1}^k \lambda_i^2$.
Also, do we have the correspondence of this in a vector form?
We have \begin{align*} k^{-1}\|A\|_F^2 = k^{-1}\sum_{i=1}^{k}\lambda_i^2 \ge \left(k^{-1}\sum_{i=1}^{k}|\lambda_i| \right)^2 \end{align*} by the inequality between quadratic means and arithmetic means. Another proof would be, with $\mathbf{\lambda} = (|\lambda_1|, \cdots, |\lambda_k|)$ and $\mathbf{1} = (1, \cdots, 1)$, we have \begin{align*} \|A\|_F^2 = k^{-1}(\mathbf{\lambda}^\intercal\mathbf{\lambda})(\mathbf{1}^\intercal\mathbf{1}) \ge k^{-1} (\mathbf{\lambda}^\intercal \mathbf{1})^2 = k^{-1}\left(\sum_{i=1}^{k}|\lambda_i|\right)^2 \end{align*} by Cauchy-Schwarz.