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

Question

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

score 1 · Accepted Answer · answered Nov 12 '13 at 22:53

1

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

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

answered Nov 12 '13 at 22:53

Vedran Šego

11,372

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 01:32

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

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