# null space

• Jan 21st 2011, 06:10 AM
Sambit
null space
Let $\displaystyle A=\begin{bmatrix} 1 & 2 & 3 & 6\\ 2 & 6 & 9 & 18\\ 1 & 2 & 6 & 12 \end{bmatrix}$.

Find a basis for the null space of A.

I got the answer $\displaystyle \begin{bmatrix} 0\\ 0\\ -2\\ 1 \end{bmatrix}$

Is it correct?
• Jan 21st 2011, 06:24 AM
TheEmptySet
Yes it is correct. By the way the matrix has Rank 3 so the dim of the nullspace must be 1 and you found 1 vector in the null space.
• Jan 21st 2011, 07:56 AM
HallsofIvy
So that, strictly speaking, you have NOT yet solved the problem. The "null space" is a one-dimensional subspace of $\displaystyle R^4$ and you have found one vector in that space. How would you represent any vector in the null space?
• Jan 22nd 2011, 05:55 AM
Sambit
any vector in the null space.....do you mean $\displaystyle \begin{bmatrix} 0\\ 0\\ -2t\\ t \end{bmatrix}$ where $\displaystyle t$ is a real number?
• Jan 22nd 2011, 07:08 AM
HallsofIvy
Yes, or, to phrase it slightly differently, $\displaystyle t\begin{bmatrix}0 \\ 0 \\ -2 \\ 1\end{bmatrix}$, the set of all real multiples of your original vector= the subspace spanned by that vector.(Clapping)