When we consider the square matrix ${\bf A}\in\mathbb{C}^{N\times N}$, then following ineqaulity always holds: $$ \left\|{\bf A^{\sf H}}{\bf A}\right\|_F = \left\|{\bf A}{\bf A^{\sf H}}\right\|_F\le\left\|{\bf A}\right\|_F^2 $$ I'm curious about when the eqaulity holds. I guess only the matrix ${\bf A}$ is zero matrix, i.e., ${\bf A}={\bf 0}$.
Is it true? or not.
By the singular value decomposition we have $$Ax=\sum_{k=1}^n \lambda_k \langle x, u_k\rangle v_k,$$ where $\{u_k\}_{k=1}^n$ and $\{u_k\}_{k=1}^n$ are orthonormal bases and $\lambda_k$ form a nonincreasing collection of nonnegative numbers. Then $$A^HAx=\sum_{k=1}^n \lambda_k^2 \langle x, u_k\rangle u_k\quad{\rm and}\quad \|A^HA\|_F=\left (\sum_{k=1}^n\lambda_k^4\right )^{1/2}$$ Furthermore $Au_i=\lambda_iv_i$ and $$\|A\|_F^2=\sum_{i,j=1}^n|\langle Au_i,u_j\rangle |^2=\sum_{i,j=1}^n\lambda_i^2|\langle v_i,u_j\rangle |^2=\sum_{i=1}^n\lambda_i^2\sum_{j=1}^n|\langle v_i,u_j\rangle |^2=\sum_{i=1}^n\lambda_i^2$$ Therefore equality holds if and only if $$\left (\sum_{i=1}^n\lambda_i^2\right )^2=\sum_{i=1}^n\lambda_i^4$$ Thus $\lambda_2=\ldots =\lambda_n=0.$
Summarizing the equality holds if and only if the rank of the matrix $A$ is less than or equal to $1.$