I am struggling with this derivation (about wireless channel capacity) while reading a textbook. It seems simple, but I couldn't figure it out. So I am very appreciative if someone explains it to me.
Let $\mathbf{H} \in \mathbb{C}^{m \times n} (m < n, rank(\mathbf{H}) = d)$ can be decomposed by SVD as $\mathbf{H} = \mathbf{U}\mathbf{\Lambda}\mathbf{V}^H$, where $\mathbf{U} \in \mathbb{C}^{m \times d}$, $\mathbf{\Lambda} \in \mathbb{C}^{d \times d}$, $\mathbf{V} \in \mathbb{C}^{n \times d}$, and $\mathbf{A} \in \mathbb{C}^{n \times r}$. Then, $$\max_{\mathbf{A}} \log_2(|\mathbf{I}+\mathbf{H}\mathbf{A}\mathbf{A}^H\mathbf{H}^H|) = \max_{\mathbf{A}} \log_2(|\mathbf{I} +\mathbf{\Lambda}^2 \mathbf{V}^H\mathbf{A}\mathbf{A}^H\mathbf{V}|),$$ where $|\mathbf{X}|$ is the determinant of matrix $\mathbf{X}$ and $\mathbf{I}$ is the identity matrix. I don't understand how we can remove $\mathbf{U}$ and $\mathbf{U}^H$ when we substitute the SVD of $\mathbf{H}$ onto the formula.
By definition of the Singular Value Decomposition, $\mathbf{U}$ is a unitary matrix, so $$\mathbf{U}^H\mathbf{U} = \mathbf{I}.$$ Therefore, $$\mathbf{H}\mathbf{A}\mathbf{A}^H\mathbf{H}^H = \mathbf{U}\mathbf{\Lambda}\mathbf{V}^H \mathbf{A}\mathbf{A}^H(\mathbf{U}\mathbf{\Lambda}\mathbf{V}^H)^H = \mathbf{U}\mathbf{\Lambda}\mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}\mathbf{\Lambda}^H\mathbf{U}^H.$$
Now, using the Weinstein–Aronszajn identity that @obareey noted in the comments, \begin{equation*} \begin{split} \det[\mathbf{I}_m + (\mathbf{U}\mathbf{\Lambda})(\mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}\mathbf{\Lambda}^H\mathbf{U}^H)] & = \det[\mathbf{I}_n + (\mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}\mathbf{\Lambda}^H\mathbf{U}^H)(\mathbf{U}\mathbf{\Lambda})] \\ & = \det[\mathbf{I}_n + \mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}\mathbf{\Lambda}^H(\mathbf{U}^H\mathbf{U})\mathbf{\Lambda}] \\ & = \det[\mathbf{I}_n + \mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}\mathbf{\Lambda}^H\mathbf{\Lambda}] \\ & = \det[\mathbf{I}_n + \mathbf{\Lambda}^2\mathbf{V}^H \mathbf{A}\mathbf{A}^H\mathbf{V}], \end{split} \end{equation*} since $\mathbf{\Lambda}$ is a square diagonal matrix.