Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a matrix A that is like this: Matrix

The question is: Find a set of 5x1 matrices whose linear span is the null space of A.

I did Guass-Jordan and I got the matrix down to: REF

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

share|improve this question
    
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
show 1 more comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.