# Range space $\mathcal{R}(\textbf{A})$ the same as $\mathcal{R}(\textbf{AA}^H)$?

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?

• Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially) – TrostAft Dec 5 '18 at 20:17
• Also shouldnt $AA^H = U \Sigma V^H V \Sigma U^H = U \Sigma^2 U^H$? Then using this it's clear that they have the same rank. – TrostAft Dec 5 '18 at 20:20
• Ah, the first resource was very helpful, I forgot to search for the question using transpose instead of conjugate transpose (to include real vector spaces and find an analog). Your second answer I think is what I was looking for - sometimes I miss the obvious. Thank you so much! – W. MacTurk Dec 5 '18 at 20:26
• @TrostAft In fact $\Sigma^2$ corresponds to $\Sigma \Sigma^H$. $\Sigma$ is not a square matrix. I agree this notation is often used, but can be misleading in some occasions. In particular, we don't get the same $\Sigma^2$ when calculating $AA^H$ or $A^HA$ – Damien Dec 5 '18 at 20:42

$$A = \sum_{i=1}^{rank} \lambda_i u_i v^H_i$$
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
• Yes: the set generated by the $u_i$ for which the $\lambda_i$ are greater than 0 – Damien Dec 5 '18 at 20:32