How check for subspaces in Linear Algebra?

[saintdick]

Homework Statement

consider R^4. Let V be the set of vectors in the form ( 2x+3y, x, 0 , -x+2x) is this a subspace

of R^4 and why? find a basis if it's possible

Homework Equations

The Attempt at a Solution

I know that the set must work under scalar multiplication and vector addition but i am confused here because of the variables. how do you actually test that? do you just choose random values for x and y?

for the second part, i know that a basis could be found if it's a subspace but i am not sure how to find it?

 

[HallsofIvy]

IF you believe 'closure' of addition and scalar multiplication is false, yes, you could look for values of x and y which make the statement untrue. That is, look for a "counter-example".

However, if it is really a subspace, closure of addition and scalar multiplication must be true for all x and no matter examples you tried, that would never prove it true for all.

Instead use algebraic theorems to show the set is closed under addition and scalar multiplication. For this paraticular example, two "generic" members of the set would be of the form (2x+ 3y, x, 0, -x+ 2y) and (2x'+ 3y', x', 0, -x'+ 2y'). The sum of those is ((2x+ 3y)+(2x'+ 3y'), x+ x', 0+ 0, (-x+ 2y)+ (-x'+ 2y'))= (2(x+x')+ 3(y+ y'), x+x', 0, -(x+ x')+ 2(y+ y')). Do you see that this is in the same subset? It must be of the form (2x"+ 3y", x", 0, -x"+ 2y"). What are x" and y" in this case?

To see that it is closed under scalar multiplication, take a general member of the set, (2x+ 3y, x, 0, -x+ 2y) and multiply by the scalar "r":
(r(2x+ 3y), rx, r0, r(-x+ 2y))= (2(rx)+ 3(ry), rx, 0, -(rx)+ 2(ry)).
Again, do you see that is of the form (2x"+ 3y", x", 0, -x"+ 2y")? What are x" and y" here?

To find a basis, note that any member of this set is of the form (2x+ 3y, x, 0, -x+ 2y)= (2x, x, 0, -x)+ (3y, 0, 0, 2y)= x(2, 1, 0, -1)+ y(3, 0, 0, 2).

 

[saintdick]

Thanks for your reply, but i am still confused? what to do you mean when you ask " What are x" and y" in this case". So let me get this straight, when you prove it the way you did, as long as the form doesn't change then that means it's a subset. For the basis, are both of those columns form one basis or do each one form a single basis?