1. Aug 18, 2011

### ohyeahstar

Just a small question, I think I may have missed this part out in our lectures or something. :|

Suppose I have a singular matrix A; will there always exist another matrix B such that AB (/BA) will be the zero matrix?

2. Aug 18, 2011

### micromass

Yes, take B the zero matrix but you mean non-zero matrices, I suppose.

Well, if A is singular, then there always exists a nonzero column vector x such that Ax=0. Then B=(x x ... x) should do the trick.

3. Aug 18, 2011

### I like Serena

Welcome to PF ohyeahstar!

Yep, as mm said!

More specifically, a matrix has a so called "null space" (or "kernel space") and a so called "column space" (or "image" or "range" of the matrix).
Any matrix with columns selected from the null space will satisfy your criterion for B.
Furthermore the columns from your matrix A "span" the column space.