# Matrix column and row space basis

I am stuck at this matrix: $\begin{pmatrix} 1 & 1 & 0 & 1 \\ -1 & 2 & 1 & 1 \\ -1 & 8 & 3 & 5 \end{pmatrix}.$ When going to normal form, we find this: $\begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$

Now it's obvious that the row space has the following basis: $\langle (1, 1, 0, 1), (0, 3, 1, 2) \rangle$

But how to determine the column space basis? I know by a theorem it must have dimension $2$.

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See here for an example of the pmatrix environment. –  Dylan Moreland Jan 21 '12 at 18:45

Here's one way: Identify the columns of your fully reduced form that contain a non-zero leading row entry (the maroon entries in the reduced form below). For your reduced form, shown below, the leading non-zero row entries are colored maroon: $$\left[ \matrix{ \color{maroon}1 & 1& 0 & 1\cr 0& \color{maroon}3& 1 & 2\cr 0& 0 &0 & 0}\right]$$ The corresponding columns are columns one and two.