Thread: 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}

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 " "?

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, " ", and then take out the variables and you will have yous spanning set. So here " " would become " " 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.

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.

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."

So basis

Originally Posted by math2009

So basis

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! .

Originally Posted by math2009

So basis

Sorry, it should be .
It's obvious, vectors are linear independent.

Originally Posted by math2009

Sorry, it should be .
It's obvious, vectors are linear independent.

Well yeah, but you have to say it!