In the paper "On best approximate solutions of linear matrix equations", there is a very small equivalence I don't know where it comes from.
Let $A$ be a matrix (either real or complex), and $\|A\|$ be "the sum of the squares of the moduli of the elements of $A$" (sic; I believe it's the Euclidean norm of $A$), and $tr(\cdot)$ is the trace operator. I want to prove that $\|A\| = tr(A^{*}A)$.
I attempted to check this with $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix}$, but $\|A\| = 30$ and $tr(A*A) = 29$.
What am I doing wrong? I think $*$ is the matrix multiplication operator, but it could also be the generalized inverse.
It's not $A*A$, it's $A^*A$ where $A^*$ is the transpose of $A$. If you calculate $\operatorname{tr}(A^*A)$ (called the Frobenius norm$^\dagger$) you'll find that it equals $\|A\|$.
$^\dagger$: Technically this is the square of the Frobenius norm.