I know this question has been asked and answered here: Can a matrix have a null space that is equal to its column space?. However, I'm not clear on the mechanism used to find an actual matrix so I figured I would make a new question. If this is inappropriate please flag or delete.
What I have done is let $A$ be a 4 x 4 matrix where $Null(A)=Col(A)$ (null space of A=Col Space of A). Therefore, the span of the columns of A = the null space of A.
I know that the RREF of A will have the bottom two rows zeroed out. But, what I don't know is where to go from here in terms of finding a matrix that satisfies the above condition.
Any help provided would be much appreciated. I'm taking an online, distance linear algebra course for credit and the materials provided are very minimal.
Best Answer
You need $A \ne 0$ such that $A^2 = 0$ and $\operatorname{rank} A = n/2$.
So, for any nonsingular $S$, you can define $A = S^{-1} J S$, where $J$ is a Jordan matrix of the form
$$J = \bigoplus_{k=1}^{n/2} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$