# Pseudoinverse of a $m \times n$ matrix

Suppose matrix $A$ has linearly independent columns. Show that $ABA=A$ and that $AB$ is symmetric where $B$ is the pseudoinverse of $A$.($B=(A^TA)^{-1}A^T$)

Is the condition that $A$ has linearly independent columns necessary?

## 1 Answer

Yes, the condition is definitely required. For example, with $A = \mathbf 0$, the zero matrix, the expression $(A^TA) = \mathbf 0$ and hence its inverse does not exist.

In fact, the expression $(A^TA)^{-1}$ is well defined at least if $A$ has linearly independent columns. This is not difficult to see, and I leave it as an exercise.

For general $A$, the Moore Penrose pseudoinverse stipulates both that $B$ must satisfy $ABA = A$ and $AB$ must be symmetric. That is, we can't prove this for general $A$, but any candidate for a pseudoinverse must satisfy at least these two conditions (along with some others). The exercise that you have is motivation for the general definition of the Moore-Penrose pseudoinverse.