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

\begin{equation} A = \pmatrix{ 1 & 2 & -3 & 1 & 5 \\ 1 & 3 & -1 & 4 & -2 \\ 1 & 1 & -5 & -2 & 12 \\ 1 & 4 & 1 & 7 & -7 } \end{equation}

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:

\begin{equation} \pmatrix{ 1 & 2 & -3 & 1 & 5 \\ 0 & -1 & -2 & -3 & -7 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -2 } \end{equation}

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.2012-09-26
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    @EuYu So from my reduced matrix, how do I form the linear span exactly? Not really sure I'm getting it2012-09-26
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    Can you identify which columns are the free variables? How do the pivot columns depend on the free variables?2012-09-26
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    @EuYu Ohhh i think I got what you meant. Can you check to see if this is correct? http://i.imgur.com/SM4iU.jpg2012-09-26
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    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 $x_4$.2012-09-26
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    @EuYu Not sure what you mean. The matrix $[1 \,\, 1]$ has non-trivial nullspace, but $x_1+x_2=b$ sure sounds consistent to me.2012-09-26

2 Answers 2

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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.

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The software Mathematica can find a null-space spanning set for Matrices given with exact coefficients:

NullSpace[{{1, 2, -3, 1, 5}, {1, 3, -1, 4, -2}, {1, 1, -5, -2, 12}, {1, 4, 1, 7, -7}}] 

gives

{{5, -3, 0, 1, 0}, {7, -2, 1, 0, 0}} 

as in copper.hat's answer. I suggest you start here if you are interested in how this is done computationally.

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    xilo, please check the dates on questions before you provide answers to them, especially if there are already other answers available. This question is over four years old! Answering old questions moves them onto the main page, which bumps newer, unanswered questions into areas of the site with less visibility.2016-10-04