Express a sum of eigenvalues ​of a matrix $ A $ in terms of $ a_ {i, j} $

Question

I have a problem that linear algebra that I can't verify:

Let $A\in\mathcal{M}_{i,j}(\mathbb{K})$ a matrix, I want to prove that

$$ \text{Tr}(AA^{t}) = \sum_{i=1}^{n}\sum_{k=1}^{m} |a_{i,j}|^{2}, $$ and, if $n=m$ (this last hypothesis may not be necessary): $$ \text{Tr}(AA^{t}) = \sum_{i=1}^{n}\sum_{k=1}^{m} |a_{i,j}|^{2} = \sum_{i=1}^{n} \lambda_{i}^{2} $$ whit $(\lambda_i)_{i=1}^{n}$ are eigenvalues.

The firt equality is fulfilled due to: \begin{equation*} (AA^{t})_{i,j} = \sum_{k=1}^{m} a_{ik}a_{kj}. \end{equation*} then \begin{equation*} \text{Tr}(AA^{t}) = \sum_{i=1}^{n} (AA^{t})_{i,i} = \sum_{i=1}^{n}\sum_{k=1}^{m} a_{ik}a_{ki} = \sum_{i=1}^{n}\sum_{k=1}^{m} |a_{ik}|^{2}. \end{equation*}

But the second equality I can't prove it. Any ideas?

Answer 1

Hint: The value of the trace shouldn't depend on the basis used. What is obtained by using the spectral basis instead of the canonical basis?