Q:

Proof for hermitian matrices

Proof$ \sum_{i=1}^{n} \lambda_i^2$ =$\sum_{i,j=1}^{n} |a_i,_j|^2$ if A=[$a_i,_j$] is hermitian and $\lambda_i$ are eigenvalues.
Been thinking about this question for a while but  I don't know where to start. I have had a linear algebra course but didn't do too well, its all very abstract for me. My biggest problem is that I do not know how to approach the questions.

A:

As hinted in the comments, consider the matrix $A^*A$. Let us calculate its trace:
$tr(A^*A) = \sum_{i=1}^n \sum_{j=1}^n ((A)_{i,j}(A^*)_{j,i}) = \sum_{i=1}^n \sum_{j=1}^n a_{i,j} * \overline {a_{i,j}} = \sum_{i=1}^n \sum_{j=1}^n |a_{i,j}|^2 = \sum_{i,j=1}^n |a_{i,j}|^2$
Since A is hermitian, we know that $A=A^*$ and therefore $A^*A=A^2$, so we get that $tr(A^2)=\sum_{i,j=1}^n |a_{i,j}|^2$
At last, we'll remember that the trace of a matrix is the sum of its eigenvalues, and the eigenvalues of $A^2$ are exactly $\{\lambda_i^2|1 \le i \le n \}$ (it is easy to be convinced so), and we get the desired identity:
$\sum_{i,j=1}^n |a_{i,j}|^2 = \sum_{i=1}^n \lambda_i^2 = tr(A^2)$

