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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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    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.)2012-11-30

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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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    Ok I see your point.2012-11-30
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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 $\begin{pmatrix} -y & -x z \\ x & -y z \\ 0 & x^2+y^2 \end{pmatrix}$