Question

For each of the following, either use the subspace test to show that the given
subset, W, is a subspace of V , or explain why the given subset is not a subspace
of V .
(b) V = lR^3 and W ={ (x, y, z) in V l x(y -z) = 0}.

Answer 1

It isn't a subspace.
Here's a hint:
Look for two elements of W (which would also be elements of IR³)
Such that if you add them, the sum is not in W.

Answer 2

A stronger hint :D
What does it mean, if
x(y - z) = 0
This is the product of a pair of real numbers, so it can only mean that either
x = 0
or
(y - z) = 0

So, take both cases, so to speak...
(If all that made sense to you :D )

Answer 3

All this says
ONE
If x = 0, it doesn't matter what y and z are, (x , y , z) ∈ W

TWO
If y = z, it doesn't matter what x is, (x , y , z) ∈ W

So take an element that fits the first case, and take another element that fits the other...

Answer 4

Ok...
(0 , 5 , 3) ∈ W
Since
0(5 - 3) = 0

(2 , 2 , 2) ∈ W
Since
2(2 - 2) = 0

But
(0 , 5 , 3) + (2 , 2 , 2) = (2 , 7 , 5)
2(7 - 5) = 4 ≠ 0
Hence
(0 , 5 , 3) + (2 , 2 , 2) = (2 , 7 , 5) ∉ W

And therefore, W is not closed under vector addition; It cannot be a vector space; it cannot be a subspace of anything.
//

Answer 5

so a counter example would be say let v=(0,1,2), u=(1,1,1), thus u+v=(1,2,3) and hence x(y-z) not = 0. Would that be enough to explain its not a subspace?

Answer 6

Well, yes, it is. Something that is not a vector space in the first place, it cannot be a subspace :)

Answer 7

haha yh you beat me to it. can you help me with the same question but about polynomials?

Answer 8

I'll do what I can.

Answer 9

ok thanks let me type it.

Answer 10

V=P_2(lR)(real polynomial functions of degree at most 2) and W={f inV lf(2)=2f(1)}

Answer 11

W={f inV l f(2)=2f(1)} *

Answer 12

Notation's getting me dizzy... hang on, processing request...

Answer 13

And its using the same question as above either use subspace test or explain why its not.

Answer 14

Okay. Do you know the subspace test?

Answer 15

Yup. Test for zero, closure under addition and closure under scalar multiplication right?

Answer 16

That's right. Now, is the 0-vector (in this case, the zero polynomial) in W?

Answer 17

Yh i think.

Answer 18

I don't think instructors will accept that :P
Let z(x) be the zero polynomial. Then z(x) is in V.
Then z(x) = 0, for all x.
Then z(2) = 0 = 2(0) = 2[z(1)]

There, we showed that the zero polynomial is in W.
Care to do the rest of the subspace test? :)

Answer 19

Ok, here's the deal. We have to check for closure under addition and closure under scalar multiplication, right? I'll do one of them, you do the other. Which of them do you want me to do?

Answer 20

you do addition, or do the start and i will try and finish it.

Answer 21

see here's how i think it works

Answer 22

Ok. So, suppose p(x) and q(x) were both in W. Then let
r(x) = p(x) + q(x)
So that:
r(2) = p(2) + q(2)
r(1) = p(1) + q(1)

Then...

r(2) = p(2) + q(2)
= 2p(1) + 2q(1) // since p and q are both in W. DON'T FORGET THAT FACT
= 2[p(1) + q(1)] // Just "undistributing the 2
= 2r(1)

And thus we've shown that r(x) which happens to be the sum of two arbitrary elements of W, is also in W, as it fits the necessary conditions to be in W.
Hence, W is closed under vector addition.

Now it's your turn :)

Answer 23

Ok. So suppose c in lR, and u(x) in W.

Im stuck. :/ what gets multiplied.

Answer 24

So suppose c in lR, and u(x) in W. Then let
r(x)=cu(x)

then
cu(2)= c*u(2) = c* 2u(1) = 2cu(1).

Thus as
r(x) = cu(x),

==> r(2)=2r(1).

and thus closure under scalar multiplication.

Answer 25

Is this correct?

Answer 26

Sorry, dozed off, big time o.o. Yep, this is correct