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?

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- Jan 21st 2011, 06:10 AMSambitnull 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 AMTheEmptySet
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 AMHallsofIvy
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 AMSambit
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 AMHallsofIvy
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)