Frobenius norm of a matrix

Question

I know that Frobenius norm of a matrix A is equal to the square root of the trace of (A*conjugate transpose(A)). But how do I prove it mathematically?

Answer 1

Presumably, you have the entry-wise definition, $$ \|A\|_F^2 = \sum_{i,j} |a_{ij}|^2 $$ In order to get to this from the trace of the product $AA^*$, use the summation definition of matrix multiplication: $$ [AB](i,j) = \sum_{k=1}^n a_{ik}b_{kj} $$ and of the trace: $$ \text{trace}(A) = \sum_{\ell = 1}^n a_{\ell \ell} $$