Show: $\sum_{i,j=1}^n |a_{ij}|^2=trA^*A=\sum^n_{i=1}\sigma_i^2$

Question

I'm trying to prove the above fact for an arbitrary matrix $A$, with eigenvalue $\lambda_i$, and singular values $\sigma_i$.

My approach so far: the trace of a matrix is the sum of its eigenvalues, and $A^*$ has the same eigenvalues as A. Diagonalize $A$, and I get the second half of the equality, since the diagonal entries of $A^*A$ are now the squares of the singular values. The first half eludes me, could someone point me in the right direction?

Answer 1

For the first half, just use the literal meaning of the trace. What are the diagonal entries of $A^*A$?

Answer 2

Let $A=(a_{ij})_{i,j}\in\mathbb{C}^{n\times n}$.

For the fist equality, just write everything out: if $(b_{ij})_{i,j}:=A^*A$, then $b_{ij}=\sum_{k=1}^n\overline{a_{ki}}a_{kj}$, so: \begin{align*} \text{tr}(A^*A)&=\sum_{i=1}^nb_{ii}\\ &=\sum_{i=1}^n\sum_{k=1}^n\underbrace{\overline{a_{ki}}a_{ki}}_{|a_{ki}|^2}\\ &=\sum_{i,k}|a_{ik}|^2 \end{align*} For the second, be careful: $A$ is not necessarily diagonalizable. However, since $A^*A$ is self-adjoint, it is diagonalizable and, furthermore, all its eigenvalues are real, namely $\sigma_1^2,...,\sigma_n^2$. In particular: $$\text{tr}(A^*A)=\sum_{i=1}^n\sigma^2_i$$