Find a basis for the subspace R3 defined by

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

- Oct 27th 2009, 09:29 AMpasleycakesFind 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 AMSwlabr
What is "$\displaystyle s$"?

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, "$\displaystyle (0, 5b, -b)$", and then take out the variables and you will have yous spanning set. So here "$\displaystyle (0, 5b, -b)$" would become "$\displaystyle (0, 5, -1)$" 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 AMHallsofIvy
- Oct 28th 2009, 07:13 AMSwlabr
- Oct 28th 2009, 04:35 PMmath2009
$\displaystyle \left[\begin{array}{c} 3a \\ 5b \\2a-b \end{array}\right]=\left[\begin{array}{c} 3a \\ 0 \\2a \end{array}\right]+\left[\begin{array}{c} 0 \\ 5b \\-b \end{array}\right]=a\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right]+b\left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right]$

So basis $\displaystyle \mathfrak{B}=span(\left[\begin{array}{c} 3 \\ 0 \\2 \end{array}\right], \left[\begin{array}{c} 0 \\ 5 \\-1 \end{array}\right])$ - Oct 28th 2009, 11:43 PMSwlabr
- Oct 29th 2009, 04:00 PMmath2009
- Oct 30th 2009, 12:43 AMSwlabr