# Constructing matrix with nullspace containing particular vector

How can we construct a matrix $A$ such that $Null(A)$ contains the vector $u=\begin{pmatrix}2\\1\\2\end{pmatrix}$?

Thanks :)

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Are you familiar with the definition of $Null(A)$ (otherwise known as $ker(A)$)? –  andybenji Nov 30 '12 at 1:32
@andybenji I know what is $Ker$ but what is $Null$ I don't know. Thanks :) –  Iuli Nov 30 '12 at 1:41
Just let $A$ be the 3x3 zero matrix. Then every vector is in $Null(A)$. Did you mean to say find all matrices $A$ with this property? –  Logan Nov 30 '12 at 1:42
The null space is the kernel. Try choosing a vector $x$ such that $x^T u = 0$. Pick $x_1=x_2 =1$ and then figure out what $x_3$ must be in order to satisfy the requirement. Then let $A = x^T$. (Logan's answer is even simpler.) –  copper.hat Nov 30 '12 at 1:49

Hint: what can you say about the rows of such a matrix?

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If $A$ is a $3 \times 3$ matrix, then find any vector $v$ that is orthogonal to $u$, and let $A=vv^T$.

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If A is 3x3 then ${\rm null}(A)$ is empty. A needs to be 2x3. –  ja72 Nov 30 '12 at 3:56
@ja72 how about $A=\begin{bmatrix}1 & 0 & -1 \\ 0 & 0& 0\\ -1 & 0 &1 \end{bmatrix}$ –  chaohuang Nov 30 '12 at 4:21
Ok I see your point. –  ja72 Nov 30 '12 at 17:21

In MATLAB

v = [2;1;2];

A = null(v.')

A = |-1/2, -1|
|   1,  0|
|   0,  1|

null(A.')= | 1 |  % which is co linear with v
|1/2|
| 1 |


in general for the ${\rm null} \begin{pmatrix} x \\ y \\ z \end{pmatrix}^\top$ I calculate $\begin{pmatrix} -y & -x z \\ x & -y z \\ 0 & x^2+y^2 \end{pmatrix}$

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