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I tried to solve this problem in the following way, I just need you to tell me if it's right.

Find a matrix $A$ such that $u$ is in $\operatorname{Null}(A)$. $$u = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$$

I found $A$ in the following way:
$$A=\begin{bmatrix} a&\frac{1}{2}u&u \end{bmatrix}$$ where $a = \begin{bmatrix} -2 \\ -3 \\ 3 \end{bmatrix}$
$$A = \begin{bmatrix} -2&1&2 \\ -3&1/2&1 \\ 3&1&2 \end{bmatrix}$$
If I reduce it to reduced row echelon form, $\dim(\operatorname{Null}(A))=1$. Is this the correct solution?

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Did you check that $Au=0$? It doesn't seem to be. [The snarky answer to this problem, as written, is $A = 0$, but maybe you want $u$ to span the null space.] – Dylan Moreland Jan 19 '12 at 15:33
You can check whether your solution is correct, since $u \in Null(A)$ if and only if $Au = 0$. Does $Au = 0$ for your choice of $A$? Also, since the problem doesn't specify you need to use a $3$-by-$3$ matrix, did you consider trying the smallest possible size for $A$? It's always a good practice to seek the simplest examples possible. Does $A$ have to have $3$ rows? Does $A$ have to have $3$ columns? – Michael Joyce Jan 19 '12 at 15:36
Is $A=0$ allowed? :-) – lhf Jan 19 '12 at 15:39
@Andrew: It seems like you might be confusing $Null(A)$ (the null space of $A$) with $Col(A)$ (the column space of $A$). – Michael Joyce Jan 19 '12 at 15:48
$u$ doesn't have to appear anywhere in the matrix $A$. All you need is some matrix $A$ for which $Au = 0$. The matrix $A$ you chose above doesn't do this. – Schmitty Jan 19 '12 at 15:51
up vote 3 down vote accepted

If you check your solution, you'll see that $A{\bf u}$ is not the zero vector; so, your solution does not work (and would have been happenstance if it did).

What you could use is that if $A{\bf u}=\bf0$, then the linear combination of the columns of $A$ given by the coordinates of $\bf u$ is zero. That is you know, with ${\bf u}=\Bigl[{ \raise1pt\scriptscriptstyle 2\atop{\raise1pt 1 \atop 2}}\Bigr]$, that

$$\tag{1} 2A_1+A_2+2A_3=\bf0, $$ where $A_i$ is the $i^{\rm th}$ column of $A$.

Any matrix $A$ with the appropriate dimensions which satisfies $(1)$ will work. So, just write down a, say, $3\times3$ matrix whose second column is the $-{ 2}$ times the sum of the first and third columns.

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$\frac{-1}{2}$? Isn't it $-2$ times the sum? – Andrew Jan 19 '12 at 15:59
@Andrew eek... Yes, thank you. – David Mitra Jan 19 '12 at 16:00

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