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

A basis for the column space of the original matrix is given by the corresponding columns of the original matrix. So, here, the first two columns of the original matrix form a basis for its column space.

What's being used here is the following fact: elementary row operations on a matrix do not change the independence relations of the columns of the resulting matrix. Row operations can change the actual column space. So, once you have an echelon form, you can easily identify its independent columns (as above). Then the "same" columns of the original matrix will form a basis for its column space.

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Find the row reduced echelon form of Transpose[A] Transpose the result

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