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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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  • $\begingroup$ Are you familiar with the definition of $Null(A)$ (otherwise known as $ker(A)$)? $\endgroup$ Commented Nov 30, 2012 at 1:32
  • $\begingroup$ @andybenji I know what is $Ker$ but what is $Null$ I don't know. Thanks :) $\endgroup$
    – Iuli
    Commented Nov 30, 2012 at 1:41
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    $\begingroup$ 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? $\endgroup$ Commented Nov 30, 2012 at 1:42
  • $\begingroup$ 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.) $\endgroup$
    – copper.hat
    Commented Nov 30, 2012 at 1:49

3 Answers 3

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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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  • $\begingroup$ If A is 3x3 then ${\rm null}(A)$ is empty. A needs to be 2x3. $\endgroup$ Commented Nov 30, 2012 at 3:56
  • $\begingroup$ @ja72 how about $A=\begin{bmatrix}1 & 0 & -1 \\ 0 & 0& 0\\ -1 & 0 &1 \end{bmatrix}$ $\endgroup$
    – chaohuang
    Commented Nov 30, 2012 at 4:21
  • $\begingroup$ Ok I see your point. $\endgroup$ Commented Nov 30, 2012 at 17:21
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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 $\begin{pmatrix} -y & -x z \\ x & -y z \\ 0 & x^2+y^2 \end{pmatrix}$

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