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In pseudoinverse of a matrix, we have a special case when the columns are linearly independent. It is mentioned in that article and in other articles that

It follows that $A^+$ is then a left inverse of $A$: $A^+A=I_{n}$.

But I think it should be right inverse because:

\begin{align} (A^*A)^{-1} A^*A = I_{n} &\implies (A^*A)^{-1} A^*AA^+ = I_{n}A^+ \\ &\implies (A^*A)^{-1} A^*I_{m} = A^+ \\ &\implies (A^*A)^{-1} A^* = A^+. \end{align}

From this we see that $AA^+= I_{m}$, then $A^+$ here is right inverse, so why is it mentioned that $A^+$ is left inverse?

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  • $\begingroup$ We have $A^+ = \left(A^*A\right)^{-1}A^{*}$. Thus $A^+A = \left(A^*A\right)^{-1}A^{*}A = I_n$. So $A^+$ is a left inverse of $A$. How did you know that $AA^{+}=I_m$? This isn't the case in general actually. Is it because you thought that since $A^+ AA^+ = A^+$, we must have $AA^+ = I_m$? Note that in general if $XY = X$ ($X,Y$ matrices), it does not necessarily mean than $Y$ is an identity matrix. $\endgroup$ Mar 3 '19 at 11:59
  • $\begingroup$ @MinusOne-Twelfth thanks for replying. Actually in what you explained, you moved from something already calculated $(A^*A)^{-1} A^* = A^+$ to prove other thing. But in my case I am moving from scratch. So in my way, the first and second step there is no problem but you are saying the third step is ambiguous. Ok, so what I meant is that if A has right inverse let's call it anything $A^+$ or ... then this thing should make the product of $AA^+ =I_{m}$ I am assuming that to find the right inverse which is $A^+$ and according to this I got the same formula but as a right inverse! $\endgroup$ Mar 3 '19 at 12:16
  • $\begingroup$ OK I think I see what you did. You effectively said "assume $A$ has a right inverse $X$", and proceeded to find what $X$ would have to be (writing $X$ for the right inverse instead of $A^+$ to avoid confusion with the $\left(A^*A\right)^{-1}A^*$). The problem is, $A$ doesn't actually have a right inverse in the first place, unless $A$ has columns that form a spanning set for $\mathbb{R}^m$ (or $\mathbb{C}^m$ in the complex case). Note that this necessitates $n\ge m$. Since you were looking at the case of independent columns, we also have $n\le m$, so $n=m$ and in fact $A$ is invertible. $\endgroup$ Mar 3 '19 at 12:23
  • $\begingroup$ Ok. In case the columns are linearly dependent and same the rows, then is there a pseudoinverse for the matrix (whether left of right)? $\endgroup$ Mar 3 '19 at 12:36
  • $\begingroup$ As that Wikipedia page said, there always exists a pseudoinverse (and it is unique too). See this part of the page (the section on Existence and uniqueness). See here if you're interested in proofs. $\endgroup$ Mar 3 '19 at 12:41
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If an $m\times n$ matrix $A$ has linearly independent columns, then $m\ge n$ and the matrix has a left inverse (which is also a right inverse if $m=n$, of course).

If $m>n$, then $A$ cannot have a right inverse.

In this case the matrix $A^*A$ is invertible, because its rank is the same as the rank of $A$, so we can consider $(A^*A)^{-1}A^*$ and this is a left inverse of $A$, because $$ \bigl((A^*A)^{-1}A^*\bigr)A=(A^*A)^{-1}(A^*A)=I $$ Set $A^+=(A^*A)^{-1}A^*$ and prove it satisfies the requirements for being the pseudoinverse in this special case.

Indeed, $AA^+A=AI=A$; also $A^+AA^+=IA^+=A^+$. Next $(A^+A)^*=I^*=I=A^+A$. The last condition to verify is $(AA^+)^*=AA^+$; now $$ (AA^+)^*=(A(A^*A)^{-1}A^*)^*=A((A^*A)^{-1})^*A^* =A((A^*A)^*)^{-1}A^*=A(A^*A)^{-1}A^*=AA^+ $$

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  • $\begingroup$ Thanks. Moore-Penrose has 4 conditions for pseudoinverse. Do you know how many conditions the left/right inverses satisfy from these Moore-Penrose conditions) $\endgroup$ Mar 3 '19 at 13:26
  • $\begingroup$ @MosabShaheen I added it, it's just a simple computation. $\endgroup$
    – egreg
    Mar 3 '19 at 13:51

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