# Space Projection of singular matrix

Describe the space that $$A$$ projects onto and the space that $$A$$ projects along$$A = \begin{bmatrix} 0 & 1 \\0&1 \end{bmatrix}.$$

A matrix, $$M$$ is a perpendicular projection operator on the column space of $$M \iff MM = M$$ and $$M = M^T.$$ Clearly, $$A$$ does not satisfy this theorem. How do I describe the space that $$A$$ projects onto and the space that $$A$$ projects along? I need help.

First you need to check that $$A$$ projects, that is $$A^2=A$$. That is true.
Then the subspace it projects to is the range of $$A$$ (I hope you can find it) and the space it projects along is the nullspace of $$A$$ (you should be able to find it too).
• The second answer there is correct. The first one is totally wrong provided $n\ge 2$ and rank$(A)<n$ – JCAA Aug 30 at 20:20