# Find a matrix with the null space equal to the column space of that matrix

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.

• What did you not understand of the link you gave? The answer is in there! – kjetil b halvorsen Nov 12 '13 at 22:46
• @kjetilbhalvorsen based on my reading of link provided the matrix ends up being all zeros except for a 4 x 4 matrix with the upper rightmost value a 2 x 2 invertible matrix, is that right? – n8sty Nov 12 '13 at 22:52

## 1 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

$$J = \bigoplus_{k=1}^{n/2} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$

• I don't understand what a Jordan matrix is--not something I've been introduced to yet in my course. – n8sty Nov 12 '13 at 22:58
• You can always ask Wikipedia. Besides, the name is not important. I wrote the form of $J$: all zeroes except ones at indexes $(2k-1,2k)$ for $k=1,\dots,n/2$. – Vedran Šego Nov 13 '13 at 1:32