$A = \begin{bmatrix}1&1&1&-1&0\\1&0&1&0&1\\0&0&1&0&0\\2&0&3&0&2\end{bmatrix}$
Find a basis for the row space, column space, and null space.
I think the first two parts are easy. First;
$rref(A) = \begin{bmatrix}1&0&0&0&1\\0&1&0&-1&-1\\0&0&1&0&0\\0&0&0&0&0\end{bmatrix}$
So my row space is just the 3 pivot rows of $rref(A)$, and the column space, the corresponding columns of $A$.
I'm a little murky on the null space. I know it's the set of all vectors $x$ which satisfy $Ax = 0$. Since I have 5 columns;
$\begin{bmatrix}1&1&1&-1&0\\1&0&1&0&1\\0&0&1&0&0\\2&0&3&0&2\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = 0$
And I can find a solution, $x = (1, 0, 0, 1, -1)^t$.
However I notice that A has rank of 3, with 5 columns, and I think this means I have to find two solutions for x (due to rank-nullity theorem).
Am I on the right track, and is there a good strategy for finding these solutions easily?
Best Answer
$x$ satisfies $Ax=0$ if and only if it satisfies $rref(A) x = 0$. [You could do it your way by looking at $Ax=0$, but it is much easier if you use the RREF.]
Writing out the last expression gives you three equations. \begin{align} x_1+x_5 &= 0\\ x_2 - x_4 - x_5 &= 0\\ x_3 &= 0 \end{align} So, any solution is of the form $$\begin{bmatrix}x_1 \\ x_2 \\ 0 \\ x_1+x_2 \\ -x_1\end{bmatrix} = x_1 \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\ -1\end{bmatrix} + x_2 \begin{bmatrix}0 \\ 1 \\ 0 \\ 1 \\ 0\end{bmatrix}.$$