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:




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?

  • $\begingroup$ Is there something that breaks down in this arguement for complex? math.stackexchange.com/q/349742 (sorry I got the wrong link initially) $\endgroup$ – TrostAft Dec 5 '18 at 20:17
  • $\begingroup$ 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. $\endgroup$ – TrostAft Dec 5 '18 at 20:20
  • $\begingroup$ 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! $\endgroup$ – W. MacTurk Dec 5 '18 at 20:26
  • 1
    $\begingroup$ @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 $A^HA$ $\endgroup$ – Damien Dec 5 '18 at 20:42

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

  • $\begingroup$ But does having the same rank here indicate that A and AA^H have the same range space? $\endgroup$ – W. MacTurk Dec 5 '18 at 20:29
  • $\begingroup$ Yes: the set generated by the $u_i$ for which the $\lambda_i$ are greater than 0 $\endgroup$ – Damien Dec 5 '18 at 20:32

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