Square of spectral radius and Frobenius norm

Question

Assume matrix $A\in\mathbb{C}^{n\times n}$ and $|\lambda_1|\geq\dots\geq|\lambda_n|$ are the absolute values of its eigenvalues. I want to prove or disprove (by a counterexample) the following claim for the square of the absolute value of the largest eigenvalue $$|\lambda_1|^2\geq\frac{1}{n}\sum_{i,j}|a_{i,j}|^2$$ Perhaps this question has already been asked here but I couldn't find it.

Answer 1

Take $A= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Then $\|A\|_F = 1$, $\lambda_1 = 0$.

However, we do have $\|A\|_F^2 = \sum_k \sigma_k^2$, where $\sigma_k$ are the singular values.

Answer 2

This is true for normal matrices (i.e. matrices satisfying $A^*A = AA^*$), but it fails in general. For example: with $$ A = \pmatrix{1&t\\0&0} $$ we find that $|\lambda_1|^2 = 1$, but $$ \frac 12 \|A\|_F^2 = \frac 12(1 + |t|^2) $$ which has no upper bound (say, for $t \geq 0$).