Proof of the definition of Frobenius norm

Question

We already know from the definition of Frobenius norm that : \begin{equation} \boxed{\mathit{||A||}_{F}=\sqrt{\mathop{\mathrm{trace}}(A^*A)}.} \end{equation} But how can we prove it since the trace of a matrix $X \in \mathbb{R}^{n\times n}$ is defined to be the sum of its diagonal elements such that : $ \mathop{\mathrm{trace}}(X) = \sum_{i=1}^{n} \mathit{X}_{ii}$.

Answer 1

Let $A$ be an $m\times n$ (complex) matrix, the Frobenius norm is originally defined as: \begin{align} \left\Vert A\right\Vert_F^2 =\sum_{i,j}\vert a_{ij}\vert^2 =\sum_{i,j}a_{ij}^\ast a_{ij}. \end{align} On the other hand, observe that $A^\ast A$ is of size $n\times n$, and we can write the diagonal entries of $A^\ast A$ explicitly: \begin{align} A^\ast A=\begin{pmatrix} \sum_{i}a_{i1}^\ast a_{i1}&*&\cdots&*\\ *&\sum_{i}a_{i2}^\ast a_{i2}&\cdots&*\\ \vdots&\vdots&\ddots&\vdots\\ *&*&\cdots&\sum_{i}a_{in}^\ast a_{in} \end{pmatrix}. \end{align} This reveals that \begin{align} \operatorname{trace}(A^\ast A) =\sum_{j=1}^n\sum_{i}a_{ij}^\ast a_{ij} =\left\Vert A\right\Vert_F^2. \end{align}