Finding a smaller subset within a set

[KongMD]

Hi, everyone. I'm really at a loss as to how to solve these problems. For example, a set S of three vectors is given, and there is a supposedly a linear combination of two of the sets that equals the third set. How do I go about solving a problem like this without writing out multiples of every set and comparing them? Is there a formula? RREF returns an inconsistent matrix, when the sets are grouped together, so I see no concrete way to solve this.

I looked in my textbook (Linear Algebra: A Matrix Approach v2e by Spence, et al.) at the example on "smaller subsets", but it just shows the original set of vectors followed by 3*v₁ + 2*v₂ = v₃, with no explanation about how they obtained those numbers.EDIT: Note - the subset must be within the same span as S

 

[mathwonk]

do you know how to do gaussian elimination?

 

[KongMD]

Yes. I thought that was the way to solve the problem, at first, but the last row in my solution matrix has the co-efficient matrix equaling 1. Shouldn't it equal 0 if it's consistent, because the solution vector is all zeros, or does consistency not matter?

 

[JThompson]

Consistency does matter, as inconsistency implies there are no solutions. What you described does not sound inconsistent, though. (But that could be because it is difficult to visualize what you are describing in the last post.)
I think it would help if you posted your work.

 

[KongMD]

Ok, here's what I've done so far. There are three sets in the original question, [-1 1], [2 -2], and [1 0]. I put those in an augmented matrix like so (sorry for crummy formatting):

1 2 1 0
1 -2 0 0

Why won't this format correctly?
\left(\begin{array}{cccc}1&2&1&0\\1&-2&0&0\end{array}\right)

RREF'ing the above matrix yields this solution matrix:

1 -2 0 0
0 0 1 0

In R2, 0 cannot be equal to 1, therefore it's inconsistent.

 

[JThompson]

Your code formatted correctly for me.

\left(\begin{array}{cccc}1&2&1&0\\1&-2&0&0\end{array}\right)

1. When reducing this matrix I get

\left(\begin{array}{cccc}1&-2&0&0\\0&4&1&0\end{array}\right)

which means the solutions to

\left(\begin{array}{ccc}1&2&1\\1&-2&0\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right)

are x=2y and z=-4y.

2. The augmented matrix

\left(\begin{array}{cccc}1&-2&0&0\\0&0&1&0\end{array}\right)

is not inconsistent even if it is not the correct RREF. z=0 is the solution for the last entry.
An inconsistent matrix is one like

\left(\begin{array}{cccc}1&-2&0&0\\0&0&0&2\end{array}\right)

because the last entry is saying that part of the "solution" is x(0)+y(0)+z(0)=2, which is not possible.

Did you do column operations on your matrix, or just row operations? Column operations do change the nullspace of a matrix, which would lead incorrect results.