# Finding a spanning set for a null space

I have a matrix $A$ that is like this:

$$A = \pmatrix{ 1 & 2 & -3 & 1 & 5 \\ 1 & 3 & -1 & 4 & -2 \\ 1 & 1 & -5 & -2 & 12 \\ 1 & 4 & 1 & 7 & -7 }$$

The question is: Find a set of $5\times 1$ matrices whose linear span is the null space of $A$.

I did Gauss-Jordan and I got the matrix down to:

$$\pmatrix{ 1 & 2 & -3 & 1 & 5 \\ 0 & -1 & -2 & -3 & -7 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -2 }$$

But doesn't that mean the matrix is inconsistent therefore there is no linear span? Any ideas? Thanks

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Inconsistent matrices are precisely the matrices for which there is a non-trivial nullspace. Remember that a matrix is consistent for all coefficient vectors if and only if it has trivial nullspace. –  EuYu Sep 26 '12 at 4:26
@EuYu So from my reduced matrix, how do I form the linear span exactly? Not really sure I'm getting it –  Richard Sep 26 '12 at 4:30
Can you identify which columns are the free variables? How do the pivot columns depend on the free variables? –  EuYu Sep 26 '12 at 4:31
@EuYu Ohhh i think I got what you meant. Can you check to see if this is correct? i.imgur.com/SM4iU.jpg –  Richard Sep 26 '12 at 4:34
Not quite. Your free-variables are your columns with no leading entries. In this case, columns $3$ and $4$. Because these two variables are free, we can set them to anything, so why not take $x_3 = 1$ and $x_4 = 1$. Now when $x_3 = 1$, how is $x_1, x_2$ and $x_5$ determined? Similarly when $x_4 = 1$ how does that determine the non-free variables? Writing down those relationships should give you two vectors, one corresponding to $x_3$ and the other to $x_4$. –  EuYu Sep 26 '12 at 4:50

Your reduced matrix is correct. First you need to characterize the set of vectors $x$ that satisfy $A x = 0$. This set is called the null space or kernel, and I use the standard notation $\ker A$.
The reduction process above corresponds to pre-multiplying $A$ by an invertible matrix $G$ such that $G A = \tilde{A}$, where $\tilde{A}$ is the reduced matrix above. Since $G$ is invertible, you have $Ax = 0$ iff $\tilde{A}x = 0$, or in other words, $\ker A = \ker \tilde{A}$. So we can focus on finding $\ker \tilde{A}$, since the matrix has a nicer form.
Suppose $\tilde{A}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5} = 0$. Then you can immediately see that we must have $x_5 = 0$. The third row tells us nothing. And the first two rows can be re-written as $\pmatrix{ 1 && 2 \\ 0 && 1} \pmatrix{x_1 \\ x_2}+\pmatrix{ -3 && 1 \\ 2 && 3} \pmatrix{x_3 \\ x_4} = 0$. Since $\pmatrix{ 1 && 2 \\ 0 && 1}^{-1} = \pmatrix{ 1 && -2 \\ 0 && 1}$, we get $\pmatrix{x_1 \\ x_2} = \pmatrix{ 7 && 5 \\ -2 && -3} \pmatrix{x_3 \\ x_4}$.
This tells us that if we select $x_3, x_4$, then $x_1, x_2$ are completely determined. Thus $\tilde{A}x = 0$ iff $x_5 = 0$, $x_3, x_4$ are arbitrary, and $x_1,x_2$ given by the above formula, or in other words $\ker A = \{\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} | x_3, x_4 \text{arbitrary} \}$.
Now you want to find a more convenient way of expressing this. Note that we can write $\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} = x_3 \pmatrix{7 \\ -2 \\ 1 \\ 0 \\ 0} + x_4 \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0}$. So we can write $\ker A = \text{sp} \{\pmatrix{7 \\ -2 \\ 1 \\ 0 \\ 0}, \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0} \}$. That is, the null space of $A$ is the span of these two vectors.