Finding a matrix where the column space is a subset of the null space

Question

Let $A_{3x3}$ be a matrix $ \ne 0$ such that the column space of $A$ is a subset of the null space of $A$. I need to find $A$.

Here's my process so far:

let $v_1, v_2, v_3$ be the column vectors of $A$

$Col(A)=c_1 v_1 + c_2 v_2 + c_3 v_3$ is a subset of $Null(A)$

let $b$ be the column space of A such that $Ax=b$, then assume $b$ is also in the null space of $A$ so $Ax=0=A(Ab)$

How do I go about finding a particular matrix that satisfies this? Am I even heading on the right direction?

Try $A=\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$. Then $\ker A = \operatorname{sp} { e_1 } = {\cal R} A$. — copper.hat, Nov 12 '13 at 20:13
@copper.hat I haven't yet been introduced to kernals in my course. Is there another way to go about this? — n8sty, Nov 12 '13 at 20:37
@copper.hat can you explain what's happening in your comment above? I don't understand the 'sp' among other things. — n8sty, Nov 12 '13 at 20:43

score 0 · Accepted Answer · answered Nov 12 '13 at 20:52

Take $A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

Then $Ax=0$ iff $x_3 = 0$. It is easy to see that the null space (also known as the kernel, or $\ker A$) is given by $\ker A = \{ (x_1,x_2, 0)^T \}_{x_1,x_2 \in \mathbb{R}}$.

Then $y = Ax$ for some $x$ iff $y = (x_3,0,0)^T$ for some $x_3 \in \mathbb{R}$, so the range space is is given by ${\cal R} A = \{ (x_1,0, 0)^T \}_{x_1 \in \mathbb{R}}$.

It should be clear that ${\cal R} A \subset \ker A$.

Finding a matrix where the column space is a subset of the null space

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