How can we construct a matrix $A$ such that $Null(A)$ contains the vector $u=\begin{pmatrix}2\\1\\2\end{pmatrix}$?
Thanks :)
How can we construct a matrix $A$ such that $Null(A)$ contains the vector $u=\begin{pmatrix}2\\1\\2\end{pmatrix}$?
Thanks :)
Hint: what can you say about the rows of such a matrix?
If $A$ is a $3 \times 3$ matrix, then find any vector $v$ that is orthogonal to $u$, and let $A=vv^T$.
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 $\begin{pmatrix} -y & -x z \\ x & -y z \\ 0 & x^2+y^2 \end{pmatrix}$