Question

Linear Algebra Problem.
Please Help!

Answer 1

number 5

Answer 2

boy that's a strange looking one, eh? :) Hmm... it looks like a subspace at first glance...

Answer 3

weird notation, but I'd venture to say that yes, 5 is a subspace.

Answer 4

Yeah, because technically it fulfills all three requirements of being a subspace, but I do not know how to write it in the proper form.

Answer 5

My teacher didn't show us how to do it and then gave us this homework lol

Answer 6

Ah... so you've never had to do a subspace proof yet?

Answer 7

well not with a matrix

Answer 8

OK, well, you can think of a matrix as basically being a vector that has been squashed into a square shape... The zero "vector" in matrix talk is \[\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\] which probably makes sense anyway.

Answer 9

So, we know that the zero vector 0 = (0,0,0) is a member of W, because we can just let x1 = 0, x2 =0, and x3 =0, giving us the matrix [0 0; 0 0].

Answer 10

I don't know if you even need the matrix for this part of the proof.

Answer 11

All that is left is to show that scalar multiplies are in W, and that W is closed under addition.

Answer 12

Oh, so I would treat it like any other subspace problem where I check the three main requirements. The 0 vector thing was throwing me off at first but the vector addition and scalar multiplication part made sense.

Think you can explain 6?

Answer 13

I'm thinking no because it is not linear

Answer 14

that would be my gut instinct as well, let's see if we can come up with something that breaks it...

Answer 15

Obviously, we can't break it with the zero vector req. :)

Answer 16

0(x,x^2)=0

Answer 17

I think you're onto something, but it's "best practice" when giving a counterexample to use actual numbers and not use "x"'s in the example.

Answer 18

So, for example, observe that (1,1) is in W since (1, 1) = (1, 1^2). However, it's multiple 2(1,1) = (2,2) is not in W because (2,2) does not equal (2, 2^2) as it should. Hence, W cannot be a subspace.

Answer 19

but how does that relate to checking for the zero vector to be an element of W

Answer 20

Good question!

Answer 21

When you are trying to show that something is NOT a subspace, all you need to do is find 1 criteria that fails. You do not have to check every property to disprove it.

Answer 22

It's like trying to show someone that a car is not a horse. All you need to do is show one thing that doesn't work, like a car doesn't have a tail, but a horse does. Therefore, a car cannot possibly be a horse. I don't know if that helps or confuses things even more. :)

Answer 23

That's math logic for you. :P

Answer 24

Well the thing is my teacher told us to go through and check these specific properties, but I understand what you are saying. Technically, though, all the properties are true, right?

Answer 25

To prove and to disprove are two separate tasks.

To prove it IS a subspace: Check all 3 properties.

To prove that it is NOT a subspace: Find 1 property that fails.

Answer 26

In fact, number 6 actually DOES have the 0 vector. Indeed, (0,0) = (0,0^2) which is in W.

Answer 27

However, the fact that number 6 breaks the requirement of a subspace to have its multiples means that we are done. There is no chance that this things is a subspace. It's a standard proof technique. :)

Answer 28

hm, but a(x,x^2)=ax,x^2
how does it break it?

Answer 29

ax^2***

Answer 30

I guess that's the danger with using a's and x's which are not concrete it sort of clouds the issue. But I see what you are saying.... Let me say it this way.

Answer 31

The test for scalar multiplies says: If v is in W, then cv is in W.

Let's translate what this is saying:

v is in W means \[v = (x, x^2)\] for some x.

cv is in W means \[cv = (x', x'^2)\] for some other x'.

Answer 32

However, if we simply multiply the v (in W by assumption) we get \[cv = c(x,x^2) = (cx, cx^2)\] However, this is not good enough. Because, cx^2 does not look like (cx)^2 all squared.

Answer 33

And that is the problem. Everything in W must look like (something, something^2).

Answer 34

Oh I see!

Answer 35

It took me a while because it is so subtle. :) That's why I (and most mathematicians) prefer to use a specific case, like (1,1) to demonstrate a false conclusion.

Answer 36

It only takes one stone to bring down the giant. :)

Answer 37

brb

Answer 38

Lol my teacher hadn't talked about specific cases. He told us to just multiply the scalar a, so I see how that can make things tricky.

Thanks for the help man. You explained it very well
:)

Answer 39

no problems, teachers forget the minor details sometimes because it's often second nature for them (but not for us!). :)

Answer 40

Very true lol

Answer 41

well, not sure which side of the world you are on, but it's late here... nights! :)

Answer 42

See ya

Answer 43

Its late here too lol

Answer 44

take care!

Answer 45

You too!