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Given a matrix $A\in \mathbb{R}^{m\times n}$, the right inverse $B\in \mathbb{R}^{n\times m}$ satisfies $AB=I_{m}$.

Here is what I learned: $B$ exists iff $A$ has full row rank (?). $B$ is not unique. One possible solution is $B=A^T(AA^T)^{-1}$ if $A$ has full row rank. This is easy to see since $AB=AA^T(AA^T)^{-1}=I$ and $AA^T$ is a full rank square matrix. [Reference: wikipedia, Berkeley, MIT ]

I tried the following calculation. It seems wrong but I cannot identify the error in reasoning. Please help me check this. $$ AB=I \Rightarrow A^TAB=A^T I = A^T \Rightarrow (A^TA)^{-1}A^TAB= (A^TA)^{-1}A^T \Rightarrow B=(A^TA)^{-1}A^T $$ assuming $A$ has full column rank.

This seems wrong because: (1) This is the solution for left inverse according to MIT , rather than the right inverse. (2) I only need the assumption that $A$ has full column rank, while the existence condition for right inverse is full row rank.

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2 Answers 2

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You assumed that $B$ is a right inverse (so $A$ has full row rank) and assumed as well that $A$ has full column rank. So you have in essence assumed that $A$ is invertible to start with.

The mistake in your reasoning is that you never checked that your $B=(A^\top A)^{-1}A^\top$ is in fact a right inverse. Did you check that $A(A^\top A)^{-1}A^\top = I_m$? Indeed, the matrix $A(A^\top A)^{-1}A^\top$ gives orthogonal projection onto the column space of $A$, and so this will be a right inverse if and only if the column space is all of $\Bbb R^m$. That is, $m=n$ and $A$ is invertible.

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  • $\begingroup$ Thanks, I got it now! My reasoning starts with $AB=I$. I thought I was only using the definition with no assumptions. I realize now that the very first step can only be written under the assumption of full row rank. And yes, as you pointed out, if I further make the assumption of full column rank, $A$ is assumed to be (two-sided) invertible. That explains the solution I got for the right inverse is also the left inverse. $\endgroup$
    – July
    Dec 28, 2022 at 20:50
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The error is this. $A^T A$ is a square matrix with the same number of columns as $A$. If $A$ has full row rank but not full column rank, it can't be invertible because its rank can't be more than the rank of $A$. So there is no such thing as $(A^TA)^{-1}$ in this case.

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  • $\begingroup$ If I just assume $A$ has full column rank, then $A^T A$ is invertible. Using the above reasoning, I get a valid right inverse $B$. This means I don't even need the condition that $A$ has full row rank, which contradicts the existing statement (wiki, another post). Therefore, either my reasoning is wrong or the existing statement is wrong. $\endgroup$
    – July
    Dec 28, 2022 at 18:44

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