Question

@jvatsim
ok so x1 = 3*x2 is the next problem
so that means (3,1)^T (6,2)^T etc so nonempty

here x1=3*x2
the next step alphaX so x = (3x2, x2)^T so alpha *x
means (alpha*3*x2, alphax2)^T
and that equals alpha*3*x2 = alpha *x2
equals alpha(3*x2 = x2) so it passes and on to step 3

x + y

so here so i say x = (3x2,x2)^t
y = (3y2, y2)^t
we can still say 3x2+3y2 is = x2 + y2 ?? i think im missing something on this las step

Answer 1

@jtvatsim

Answer 2

i misspelled your name above, sorry

Answer 3

No problems. I'll be right back.

Answer 4

First part looks good for nonempty.

Answer 5

cool. the first step is the easy one

Answer 6

Yep, you can always just show that the zero vector is in the set as a backup. Your way is fine though.

Answer 7

Second part is a little convoluted, but I'll get back to that shortly, I think you are on the right track. :)

Answer 8

how would you show the first step. i thought x1 was just 3 times x2 and since its in the x y plane its non empty

Answer 9

First step I would just say (0,0) is in the set since 0 = 3*0. It's kind of "OK, duh..." but it works to prove the point.

Answer 10

oh, wow ok yeah. makes sense

Answer 11

For the second part, let's be clear first on what we will start with and what the goal is:

START: x = (x1,x2) where x1 = 3x2.
GOAL: alpha x = (alpha x1, alpha x2) where alpha x1 = 3(alpha x2).

Notice the pattern.

Answer 12

multiplying by alpha doesnt really change anything. is that what you are asking?

Answer 13

so it is still 3x2 = x2 so it would pass, right?

Answer 14

Yes, I think it is just a matter of writing style at this point. :) Here's how I would phrase the actual proof for clarity.

Answer 15

Suppose we have x = (x1,x2). Then x1 = 3*x2.

Now, alpha*x1 = alpha*3*x2, that is, alpha*x1 = 3*(alpha*x2).

So, alpha*x = (alpha*x1, alpha*x2) is in the set.

I would prefer to not convert x1 into 3*x2 in the proof. That turned me around a bit. But I see what you did. :)

Answer 16

ok. so we have the same idea for step 2, just worked it differently?

Answer 17

Wait... you say that "and that equals alpha*3*x2 = alpha *x2" but that isn't actually true.

Answer 18

ok. if i go with what i was doing thats where i get a little confused for step 3
x + y

for my way of working would be x = (3x2,x2) y = (3y2, y2)
(3x2+3y2) would be 3(x2 + y2)...... should i just do what you did and make a note taht x1 = 3x2 and just use x1 in the proof?

Answer 19

OK, let's hold off on part 2 for the time being. I want to try to do it your way for step 3. I think that might be more insightful.

Answer 20

With your way, you have x = (3x2, x2) and y = (3y2, y2). OK, that's fine.

What happens now is that we have no equation to work with. x1 = 3x2 is totally pointless right now since you already used it to convert.

So, for x + y we need to show that x+y = (x1+y1, x2+y2) satisfies (x1+y1) = 3*(x2+y2). Is that actually true?

Well you would say that x + y = (3x2 + 3y2, x2 + y2). And, yes, the first component 3x2 + 3y2 = 3*(x2 + y2) the second component.

Thus, x + y is in the set. I have to say, it's a little unorthodox, but it seems to work.

Answer 21

Your teacher is probably going to expect to see something with x1's in the proof, but I don't want to discount your way of doing it either. :)

Answer 22

ok. i think your way would be better because its shorter. and i guess easier. i thought i had to use the 3x2 because the original problem said x1 = 3x2. so i thought we had to use it throughout

Answer 23

Yeah, the equation is really just used as a "check" to make sure that the components work like they are supposed to.

Answer 24

ok. im going to redo it and just make a note that x1 = 3x2 and use x1 in the proof

Answer 25

i scrolled up and was reading and i think that is a lot less confusing than what i was trying

Answer 26

Alright, I'd be happy to check it. It's always good to experiment with your own methods first so that you can sort out the process in your head. :)

Answer 27

ok. ill be back in a few and post what i come up with

Answer 28

just a quick question. we are saying \[\alpha x=0 \] right?

Answer 29

for the second step

Answer 30

We need to show that \[\alpha x\] satisfies \[\alpha x_1 = 3*(\alpha x_2)\]

Answer 31

ok. so for step 2 (it applies to all problems depending on condition of x1 and x2) should i write \[\alpha x = 3*\alpha x2\]

Answer 32

i think i answered my own question. yes

Answer 33

before working it

Answer 34

ok so for the third step (not doing what i originally did) would i again say x1=3x2
then X = (x1,x2) Y=(y1,y2)
then X + Y = (x1+y1), (x2+y2)

but that doesnt make sense. since i am trying to show x1 = 3x2. or would that actually be the proof because i cant think of a example that would counter that. (unless there is one and i cant think of it)