Inverse of a singular Matrix

My question is why we have to restrict over selves to find inverse for non singular matrix?

We can define inverse of a square matrix as follows:

A matrix $$B$$ is said to be inverse of $$A$$ if $$BA = C$$, where $$C$$ is the matrix obtained by $$A$$ by applying row transformation (some what like normal form). Matrix $$C$$ must satisfy following properties:

1. All zero rows are at the bottom
2. leading entry of each non-zero row is $$1$$.
3. $$C$$ is the matrix obtained by applying row transformation to the maximum extent. The matrix $$C$$ is identity matrix if $$A$$ is non singular. Only in that case, $$BA = AB = I$$ holds.

In this way we can associate with every square matrix a matrix $$B$$ (inverse) and $$C$$ (some what normal form of $$A$$). This will help to solve system of equations even when they have infinite solutions! Am I right?

• Are you just thinking of the Reduced Row Echelon Form for your $C$? The decomposition that you describe is presumably similar in purpose but not as useful as the LU Decomposition.
– EuYu
Sep 17 '13 at 6:37
• By definition a square matrix is singular if it has no inverse. So whatever $B$ is when $A$ is singular, you should not call it an inverse of $A$ Sep 17 '13 at 6:45

The matrix $B$ you describe represents (by its left multiplication) a combination of row operations that will bring $A$ into into reduced row echelon form (at least I guess that is what you wanted to describe). This is indeed useful for giving the complete solution to linear systems. The main problem with this definition is that if $A$ is singular then $B$ is not unique. Indeed one can left-multiply any such matrix $B$ by any matrix whose $r$ first columns are those of the identity matrix (where $r$ is the rank of $A$), and the other columns are completely arbitrary (if you want $B$ to correspond to a combination of row operations you must ensure that $\det B\neq0$, but that is all, and this still leaves a lot of freedom when $r$ is not maximal).

• Thank you Sir. I know that B is not unique. Still i think we do bit research here. It may have lot of applications. Your answer is appreciated. Sep 18 '13 at 6:46
• One more think can be added here. We can call matrix B as pseudo-inverse of A. Sep 18 '13 at 6:52
• Calling $B$ a pseudo-inverse would be confusing, since that term is already in use, and (unlike $B$) it is well defined. Sep 18 '13 at 8:43

An inverse of a matrix is one which after matrix multiplication results in an identity matrix (I). So there is no relevance of saying a matrix to be an inverse if it will result in any normal form other than identity.

• The question does not involve an inner product sturucture; the definition linked to does. Sep 17 '13 at 6:49
• Did you carefully read the article? Sep 17 '13 at 7:04
• Well, I read it a bit rapidly, I admit. But it is full of "Hermitian", "orthogonal", and the lead says "least squares", so clearly the inner product structure is essential. Also the Moore-Penrose pseudoinverse is unique, and what OP describes is not unique. Sep 17 '13 at 7:21
• As diplomats say, your words do not correspond to reality. Sep 17 '13 at 7:55
• Maybe you are on another version of the WorldWideWeb than I am, but when I follow the link in my browser I do see the terms that I cited. So I don't know exactly which reality does not correspond to my words. Sep 17 '13 at 8:01