I'm working on a problem as follows:
Given $\textbf{A}\in\mathbb{C}^{M\times N}$, show that $\mathcal{R}(\textbf{A})=\mathcal{R}(\textbf{AA}^H)$
where $\mathcal{R}()$ denotes the range space of a matrix/transformation and $*^H$ denotes the conjugate transpose of a matrix.
I'm trying to show this by using SVD, but I get stuck relating $\textbf{A}$ to $\textbf{AA}^H$. So far, I have:
$\textbf{A}=\textbf{U}\Sigma\textbf{V}^H$
$\textbf{AA}^H=\textbf{U}\Sigma\textbf{V}^H\textbf{A}^H$
$\textbf{AA}^H=\textbf{U}\textbf{U}^H\Sigma\Sigma^H\textbf{V}^H\textbf{V}$
Note that $\textbf{U}$ is the unitary $m\times m$ matrix in SVD, $\textbf{V}^H$ is the unitary $n\times n$, and $\Sigma$ is the matrix of singular values.
I know that the columns of $\textbf{U}$ form a basis for $\mathcal{R}(\textbf{A})$, but when I use the SVD of A to try to do SVD on AA^H, I get to $\textbf{UU}^H$, which is just the identity matrix and thus can't be a basis for a complex space?
Am I going about this all wrong, or is there some misstep I've made that's preventing me from solving the problem?
Best Answer
For this kind of problem, it is convenient to use the following form of SVD:
$$A = \sum_{i=1}^{rank} \lambda_i u_i v^H_i$$
Where the $\lambda_i > 0$
Then, $$ AA^H = \sum_{i=1}^{rank} \lambda^2_i u_i u^H_i$$
It follows immediately that $A$ and $AA^H$ have the same rank