Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks :)

share|improve this question
    
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
2  
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

3 Answers 3

up vote 1 down vote accepted

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

share|improve this answer

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

share|improve this answer
    
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 $\begin{pmatrix} -y & -x z \\ x & -y z \\ 0 & x^2+y^2 \end{pmatrix}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.