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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – EuYu
    Sep 17 '13 at 6:37
  • $\begingroup$ 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 $A$ $\endgroup$ 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).

  • $\begingroup$ 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. $\endgroup$ Sep 18 '13 at 6:46
  • $\begingroup$ One more think can be added here. We can call matrix B as pseudo-inverse of A. $\endgroup$ Sep 18 '13 at 6:52
  • $\begingroup$ Calling $B$ a pseudo-inverse would be confusing, since that term is already in use, and (unlike $B$) it is well defined. $\endgroup$ 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 Moore-Penrose pseudoinverse is it.

  • $\begingroup$ The question does not involve an inner product sturucture; the definition linked to does. $\endgroup$ Sep 17 '13 at 6:49
  • $\begingroup$ Did you carefully read the article? $\endgroup$
    – user64494
    Sep 17 '13 at 7:04
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 17 '13 at 7:21
  • $\begingroup$ As diplomats say, your words do not correspond to reality. $\endgroup$
    – user64494
    Sep 17 '13 at 7:55
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 17 '13 at 8:01

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