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As we know, the left pseudo-inverse of matrix $A$ is $$A^{\dagger}_l = (A^TA)^{-1}A^T,$$ and we have $A^{\dagger}_lA = I$. The right pseudo-inverse of matrix $A$ is $$A^{\dagger}_r = A^T(AA^T)^{-1},$$ and we have $AA^{\dagger}_r = I$.

Can we assume $A^{\dagger}_l$ and $A^{\dagger}_r$ to be close enough that at some times we can exchange them, such as assuming $A^{\dagger}_rA\approx I$?

Is there any reference that discuss this issue?

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Well, you'd better be assuming that $\:A^T\hspace{0.01 in}A\:$ and $\:A\hspace{0.015 in}A^T\:$ are invertible. $\;\;$ –  Ricky Demer Sep 19 '12 at 9:41
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2 Answers

up vote 4 down vote accepted

Unless $A$ is square, with full rank, you cannot have both a left and right pseudoinverse. If they both exist, they are equal and are called the inverse, as you would usually have.

If $A$ is not square, $A\in \mathbb{C}^{m\times n}$, then $$ \operatorname{rank}(A)\leq\min(m,\,n).$$

As such, $A^TA\in\mathbb{C}^{m\times m}$ and $AA^T\in\mathbb{C}^{n\times n}$ and not both of these can be invertible.

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$\def\Mat{\mathop{\mathrm{Mat}}\nolimits}\def\R{\mathbb R}\def\l{\text{left}}\def\r{\text{right}}\def\rank{\mathop{\mathrm{rank}}\nolimits}$Let $A \in \Mat_{n,m}(K)$. If $A^\dagger_\r$ exists, then $AA^T \in \Mat_n(K)$ is invertible, so $\rank A = n$, hence $n \le m$. If $n < m$, then $A^\dagger_\r A$ has $\rank (A^\dagger_\r A) \le n < m$, so $A^\dagger_r A \not\approx \mathrm{Id}$. If $n = m$, then - as $\rank A = n$ - $A$ is invertible, so $A^\dagger_r = A^{-1}$ and - of course $A^\dagger_\r A = \mathrm{Id}$.

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