Question

Which of the following sets are subspaces of R^4? W_1 ={[x;y;z;w]: x+3y=1 and x+2z=3}

Answer 1

I know that subspaces need to be closed under addition and multiplication but I don't know how to show if this is true with this question

Answer 2

the first (and easiest) rule you want to check is if the 0 vector is in the space. If the 0 vector isnt in there, its not a subspace.

So does the vector (0,0,0,0) meet that requirement?

Answer 3

How would you check to see if there is a vector (0,0,0,0) would you put the two vectors into a matrix form and row reduce?

Answer 4

you want to put it to the test against the rule set for the space. So your rule is, a vector (x,y,z,w) is in W_1 if:
\[x+3y = 1, x+2z = 3\]

so we just want to check if (0,0,0,0) fits that bill.
it doesnt, because:
\[x+3y = 1 \iff 0+3(0) = 1 \iff 0 = 1\]
isnt true.

Answer 5

So because the zero vector isnt in the space, W_1 isnt a subspace.

Answer 6

so first you plug in the 0 into the equation and check to see if they are valid?

Answer 7

what do you after if it is valid? how would you check for the closure under addition and multiplication

Answer 8

i'll write it out on paper. one sec.

Answer 9

alright, here ya go. What you need to do to prove/disprove closure under addition is this:

if v_1 and v_2 are vectors in W_1, is (v_1+v_2) also in W_1?
when working it out, you need to use the fact that v_1 and v_2 satisfy the rule to be in the subspace.

Answer 10

Okay thanks, I learnt this like last month and forgot it so soon, thanks for the refresher.