# Find a basis for the subspace R3 defined by {(3s,5b,2a-b):a,b R}

• Oct 27th 2009, 09:29 AM
pasleycakes
Find a basis for the subspace R3 defined by {(3s,5b,2a-b):a,b R}
Find a basis for the subspace R3 defined by

{(3a,5b,2a-b): a,b are members of R}
• Oct 27th 2009, 09:54 AM
Swlabr
Quote:

Originally Posted by pasleycakes
Find a basis for the subspace R3 defined by

{(3s,5b,2a-b): a,b are members of R}

What is " $s$"?

When you are given a set in a general form you can quite easily come up with a spanning set. You just split up the general form into vectors containing one and only one of the variables. For instance, " $(0, 5b, -b)$", and then take out the variables and you will have yous spanning set. So here " $(0, 5b, -b)$" would become " $(0, 5, -1)$" which is a member of your spanning set.

You then have to ask if this spanning set is a basis or not. That is, is it linearly independent.
• Oct 28th 2009, 06:43 AM
HallsofIvy
Quote:

Originally Posted by pasleycakes
Find a basis for the subspace R3 defined by

{(3a,5b,2a-b): a,b are members of R}

So x= 3a, y= 5b, z= 2a- b.

Letting a= 1, b= 0, x= 3, y= 0, z= 2 so <3, 0, 2> is such a vector.

Letting a= 0, b= 1, x= 0, y= 5, z= -5 so <0, 1, -5> is such a vector.

I will leave it to you to show that those two vectors span the subspace and are independent.
• Oct 28th 2009, 07:13 AM
Swlabr
Quote:

Originally Posted by HallsofIvy
Letting a= 0, b= 1, x= 0, y= 5, z= -5 so <0, 1, -5> is such a vector.

This should read "Letting a= 0, b= 1, x= 0, y= 5, z= -1 so <0, 5, -1> is such a vector."
• Oct 28th 2009, 04:35 PM
math2009
$\left[\begin{array}{c} 3a \\ 5b \\2a-b \end{array}\right]=\left[\begin{array}{c} 3a \\ 0 \\2a \end{array}\right]+\left[\begin{array}{c} 0 \\ 5b \\-b \end{array}\right]=a\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right]+b\left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]$

So basis $\mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$
• Oct 28th 2009, 11:43 PM
Swlabr
Quote:

Originally Posted by math2009

So basis $\mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$

No, for two reasons. Firstly, you assume that the set is linearly independent. This needs to be proven. Secondly, the basis is not the span of those vectors - that's the whole vector space! $V=span(\mathfrak{B})$.
• Oct 29th 2009, 04:00 PM
math2009
Quote:

Originally Posted by math2009
$\left[\begin{array}{c} 3a \\ 5b \\2a-b \end{array}\right]=\left[\begin{array}{c} 3a \\ 0 \\2a \end{array}\right]+\left[\begin{array}{c} 0 \\ 5b \\-b \end{array}\right]=a\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right]+b\left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]$

So basis $\mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$

Sorry, it should be $\mathfrak{B}=(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$.
It's obvious, vectors are linear independent.
• Oct 30th 2009, 12:43 AM
Swlabr
Quote:

Originally Posted by math2009
Sorry, it should be $\mathfrak{B}=(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$.
It's obvious, vectors are linear independent.

Well yeah, but you have to say it!