# How to prove that any matrices have their own generalized inverse.

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.

• 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? Commented Aug 6, 2020 at 8:20
• @BenGrossmann Yes I do. That is exactly what I'm asking. Sorry for my English skills. Commented Aug 6, 2020 at 8:23
• Not a problem, I just wanted to check Commented Aug 6, 2020 at 8:23
• We could simply note that the Moore-Penrose pseudoinverse exists and has this property. Commented Aug 6, 2020 at 8:54

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.

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