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Let $A$ be a matrix with a form $(m.n)$, and $X$ be a matrix with a form of $(n,m)$.

If $AXA = A$, $X$ is called a generalized inverse of $A$.

How can we prove that any matrices have their own generalized inverse?

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----What I have found-----------------------------------------

When $A$ is a row echelon form matrix $F=F(m,n;r)=\begin{pmatrix} E_{ r } & O \\ O & O \end{pmatrix}$, $Y=\begin{pmatrix} E_{ r } & Y_{ 2 } \\ Y_{ 3 } & Y_{ 4 } \end{pmatrix}$ is always $F$'s generalized inverse.

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  • $\begingroup$ Your question is written a bit strangely. I think that you're asking how to show that every matrix has a generalized inverse, is that correct? $\endgroup$ Commented Aug 6, 2020 at 8:20
  • $\begingroup$ @BenGrossmann Yes I do. That is exactly what I'm asking. Sorry for my English skills. $\endgroup$ Commented Aug 6, 2020 at 8:23
  • $\begingroup$ Not a problem, I just wanted to check $\endgroup$ Commented Aug 6, 2020 at 8:23
  • $\begingroup$ We could simply note that the Moore-Penrose pseudoinverse exists and has this property. $\endgroup$ Commented Aug 6, 2020 at 8:54

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One approach is to use the fact that $A$ has a rank factorization $A = CF$ (such a factorization can be attained, for instance, using row-reduction). Because $C$ has full column-rank, it has a left-inverse $C^g$. Because $F$ has full row-rank, it has a right inverse $F^g$. If we defined $X = A^g = F^g C^g$, then we find that $$ AXA = (CF)(F^gC^g)(CF) = C(FF^g)(C^gC)F = CF = A. $$


Proof of the existence of the one-sided inverses:

Suppose that $F$ is $m \times n$ with rank $m$ ($m \leq n$). Because $F$ has rank $m$, $F$ has a set of $m$ linearly independent columns (say that the columns $i_1,\dots,i_m$ are linearly independent; we could see that such columns must exist because the reduced row echelon form exists). Let $M$ denote the matrix whose columns are $e_{i_1},\dots,e_{i_m}$, where $e_i$ is the $i$th column of the identity matrix of size $n$.

The matrix $FM$ is square and invertible, since its columns are linearly independent. The matrix $F^g = M(FM)^{-1}$ is a right-inverse of $F$, which is to say that $FF^g = I$.

For a matrix $C$ of full column-rank, we see that $C^T$ has full row-rank and therefore has right-inverse $[C^T]^g$. It follows that $$ [[[C^T]^g]^T C]^T = C^T [C^T]^g = I, $$ so that $[[C^T]^g]^T C = I$. That is, $[[C^T]^g]^T$ is a left-inverse to $C$.


Alternatively: to construct $F^g$, we could have used the fact that $FF^T$ must be invertible, from which it follows that $F^T(FF^T)^{-1}$ is a right-inverse to $F$. This coincides with the Moore-penrose pseudoinverse.

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  • $\begingroup$ Is $FC$ always an invertible matrix? $\endgroup$ Commented Aug 6, 2020 at 8:35
  • $\begingroup$ @ohisamadaigaku Good point, actually I realize now it isn't. $\endgroup$ Commented Aug 6, 2020 at 8:50
  • $\begingroup$ @ohisamadaigaku I was making things too complicated... see my latest edit. $\endgroup$ Commented Aug 6, 2020 at 8:59

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