Show that a set of vectors spans a subspace

[mattqchou]

Homework Statement

Show that {(1, 2, 3), (3, 4, 5), (4, 5, 6)} does not span R³. Show that it spans the subspace of R³ consisting of all vectors lying in the plane with the equation x - 2y + z = 0.

Homework Equations

The Attempt at a Solution

I made a matrix of:
A = [ 1 3 4 ; 2 4 5; 3 5 6] and reduced it to row-echelon form to determine rank. I found that rank(A) = 2. Therefore, it can't span R³, since rank(A) < n for Rⁿ.

But then I need to show that it spans the plane given by the equation x - 2y + z = 0.

I thought that perhaps rewriting the problem as: z = -x + 2y and plugging back in for z. But that doesn't really tell me anything.

I also tried to show the span by demonstrating where it has linear combinations by writing span(F) = span{ax - 2by + cz = 0} but I don't know what to do from here.

I was able to find the solution in the solutions manual, but I can't decipher what it's saying or how it did it. It's not using a method described in the book.

Could someone explain this better or give me another method to figure out whether a set spans a subspace?

 

[mfb]

They find explicit coefficients for the vectors (1,2,3) and (3,4,5) to express every vector in the plane as linear combination of these vectors.
The coefficients are (-2x+3y/2) and (x-y/2). Multiply them with the vectors, simplify and you get a general vector in the discussed plane.

 

[mattqchou]

They find explicit coefficients for the vectors (1,2,3) and (3,4,5) to express every vector in the plane as linear combination of these vectors.
The coefficients are (-2x+3y/2) and (x-y/2). Multiply them with the vectors, simplify and you get a general vector in the discussed plane.

Hi,

Could you explain how you find explicit coefficients? I've never heard of this method in my class before. I also can't find that in the book or the glossary.

Thanks!

 

[mfb]

See the calculation above, it does exactly this.
It starts with an arbitrary vector in the plane and assumes that it can be written as $\alpha (1,2,3) + \beta (3,4,5)$ and then solves for the two unknowns.

 

[vela]

But then I need to show that it spans the plane given by the equation x - 2y + z = 0.

I thought that perhaps rewriting the problem as: z = -x + 2y and plugging back in for z. But that doesn't really tell me anything.

Sure it does. You now have three equations that describe the vectors that lie in that plane:
\begin{align*}
x &= s \\
y &= t \\
z &= -s + 2t
\end{align*} or in vector form,
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = s\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix} + t\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}.$$ Can you take it from there?