*(Note: this is not a duplicate as suggested. I am asking for an inequality.)*

We know that for any complex square matrix $A$, we have

$$\sum_{k=1}^n a_{kk} = \sum_{k=1}^n \lambda_k.$$

I can see this relation in trace property that it is equal sum of diagonal elements.

But how to show that the sum of absolute values of the diagonal elements is less than or equal to the sum of absolute values of the eigenvalues? That is, do we have

$$\sum_{k=1}^n |a_{kk}| \le \sum_{k=1}^n |\lambda_k|\ ?$$