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So I have a matrix, $\mathcal A$, and its pseudo inverse, $\mathcal A^\dagger$. If $\mathcal A$ is $\mathcal m \times n$ with rank $\mathcal n$, what would be the rank of this new identity matrix formed by the operator $\mathcal AA^\dagger$? This identity matrix would have to be square, right? Does this mean it's in subspace $\Bbb R^n$?

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    what does "this new identity matrix formed by the operator $\mathcal A \mathcal A^\dagger$" mean? – 2017-02-08
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    I thought that a matrix times its inverse equals the identity matrix, right? – 2017-02-08
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    Non-square matrices don't have an inverse. It is called *pseudo* inverse for reason. – 2017-02-08
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    Well shoot. Then what is the significance of $\mathcal A A^\dagger$? – 2017-02-08
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    it depends on which pseudo inverse you consider. – 2017-02-08
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    If you’re talking about the Moore-Penrose pseudoinverse here, then $\mathcal A\mathcal A^\dagger$ and $\mathcal A$ have the same image. This is one of the results of [your previous question](http://math.stackexchange.com/q/2134089/265466). – 2017-02-08

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If $m \ge n$ then the rank of $A$ is less than or equal to $n$. If $m

The Moore-Penrose pseudo inverse of $A$ is given by $A^{+} = V S^{-1} U^t$. Thus, $AA^{+} = USV^t V S^{-1} U^t = U U^t$ which is an $m \times m$. This is an identity matrix if and only if $m=n$ in which case, you can replace the pseudo inverse by the standard inverse. The rank of $AA^{+}$ is given by the rank of $U$, which is $n$.

Note that $AA^{+}A = A$. If $A$ is a non-singular square matrix then $AA^{-1}A = A$. Thus, the pseudo inverse behaves like a normal inverse in this case.