Let $$A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$$ where $A_{11} \in \mathbb{C}^{k \times k}$ and $A_{22} \in \mathbb{C}^{(n-k) \times(n-k)}$ for some $1 \leq k<n$. Prove that if $A$ is normal, then $\|A_{21}\|_{\text{F}} = \|A_{12}\|_{\text{F}}$.
I tried to use block matrix multiplication or Schur decomposition, but no luck. Any suggestions?
Since $A$ is normal we have $$A^h_{11}A_{11} + A^h_{21}A_{21} = AA^h_{(k\times k)} = A_{11}A^h_{11} + A_{12}A^h_{12} $$ where $AA^h_{(k\times k)}$ is the matrix $AA^h$ with rows and columns of index $>k$ set to $0$.
Thus $$\operatorname{Tr}(A^h_{11}A_{11}) + \operatorname{Tr}(A^h_{21}A_{21}) \\= \operatorname{Tr}(A^h_{11}A_{11} + A^h_{21}A_{21}) \\= \operatorname{Tr}(A_{11}A^h_{11} + A_{12}A^h_{12}) \\= \operatorname{Tr}(A_{11}A^h_{11}) + \operatorname{Tr}(A_{12}A^h_{12}) \\=\operatorname{Tr}(A^h_{11}A_{11}) + \operatorname{Tr}(A_{12}A^h_{12})$$ where the last equality holds since $\operatorname{Tr}(CD) = \operatorname{Tr}(DC)$ for any pair of matrices $C,D$.
Thus $$\operatorname{Tr}(A^h_{21}A_{21}) = \operatorname{Tr}(A_{12}A^h_{12})$$ which is equivalent to the thesis.