Show algebraically that a vector w can be written as the sum of two other vectors?

Jormungandr

1. The problem statement, all variables and given/known data
Let v1 = <1, 1> and v2 = <-1, 1>. Show that for any vector w in the plane one can find constants c1 and c2 so that w = c1v1 + c2v2. (Hint: Express w in component form and obtain two linear equations for the unknowns c1 and c2.


2. Relevant equations



3. The attempt at a solution

Alright, so the geometric proof to this question is easy enough, but our professor wants us to find an algebraic one, and he said that it's a good example of an exam question. So I want to make sure I understand the proofs behind these questions before exam time comes.

Since w = c1v1 + c2v2, first I multiplied c1 and c2 by the vectors <1, 1> and <-1, 1> respectively:

w = c1<1, 1> + c2<-1, 1>
w = <c1, c1> + <–c2, c2>
w = <c1 – c2, c1 + c2>

Hence, I have the vector w in terms of its components. And this is where I'm stuck. I looked online at stuff about the linear dependence of vectors, and how if you equate the vector set to 0 and that is the only solution then they are linearly independent and that you can't form the third vector w from them. And in fact that's what happened when I tried to go further with that:

c1v1 + c2v2 = w = <c1 – c2, c1 + c2> = <0, 0>

=> c1 – c2 = 0, c1 = c2
and c1 + c2 = 0

=> c1 + c1 = 2c1 = 0, in which case I just get c1 = 0 and then hence c2 = 0, and since 0 is the only solution I get that they are linearly independent even though I *know* for sure that geometrically you can make them into w.

Help would be much appreciated! I've been stuck on this for a while and it's really starting to irritate me. :(

LCKurtz

You want ##\langle c_1-c_2,c_1+c_2\rangle=\langle w_1,w_2\rangle##.

Jormungandr

Hmm. I suppose that makes sense, I might have overlooked that fact. So then,

<w1, w2> = <c1 – c2, c1 + c2>
w1 = c1 – c2
w2 = c1 + c2

So the horizontal component of our vector w was shown to be c1 – c2, and the vertical component was shown to be c1 + c2. Alright, that's cool. Is that then the solution? But what does that mean? The question asks basically to show that you can add vectors multiplied by scalars to equal some third vector, right? Does describing w completely in terms of c1 and c2 mean that we've proved it, or is there some last step I'm missing? I haven't really had an "Aha!" moment here, just a little more confusion...

LCKurtz

You are given ##w_1## and ##w_2##. You have to tell what ##c_1## and ##c_2## work.

Jormungandr

I'm sorry, I'm not entirely sure I understand what you're saying here. When you say we're given ##w_1## and ##w_2##, do you mean what we have just found, i.e. ##w_1## = <##c_1## – ##c_2##>? Or do you mean ##w_1## was something that was given to us earlier in the problem?

And I'm not sure I understand what the second statement is trying to say. Could you perhaps go in a little more depth? I'm at a loss here.

LCKurtz

You were given two particular vectors v1=<1,1> and v2=<-1,1>. You are supposed to show that given any vector w = <w1,w2> you can find a linear combination of v1 and v2 that give that w. That's where your equation came from. Finding a linear combination means finding the constants c1 and c2 that work. They are the only unknowns in those equations. Solve for them.

Jormungandr

Okay, I think I might have it.

So say w = v1 + v2, in which case w = <0, 2>. Hence, w1 = 0 and w2 = 2.

So:
c1 – c2 = 0, so c1 = c2
and c1 + c2 = 2

But c1 = c2, so
c1 + c1 = 2
2c1 = 2
c1 = 1
Hence c2 = 1.

But the problem I have is that we calculated c1 and c2 to be 1, but after we assumed them to be 1 when adding v1 and v2. Is that okay in the proof? To assume some values for the components so that we can prove that they are the components later on using a derived formula?

Dick

Not really. Do the same exercise where w=<w1,w2>. You should still be able to find c1 and c2 in terms of w1 and w2.

Jormungandr

Okay... So:

<w1, w2> = <c1 – c2, c1 + c2>
w1 = c1 – c2 and w2 = c1 + c2

c1 = w1 – c2 and c1 = w2 – c2

So:
w1 + c2 = w2 – c2
w1 = w2 – 2c2
c2 = (w2 – w1)/2

Plugging this back into c1 = w2 – c2:
c1 = w2 – (w2 – w1)/2
c1 = (2w2 – w2 – w1)/2
c1 = (w2 – w1)/2

And hence c1 = c2 = (w1 – w1)/2

So we've proven that the vector w arises when c1 and c2 are equal and we add v1 and v2?

vela

That can't possibly be right. Check your algebra.

Jormungandr

You were right, there was a mistake in my signs.

c2 = (w2 – w1)/2
c1 = (w2 + w1)/2

That should be right.

And in fact, when I plug in these values for c1 and c2 in the c1<1, 1> + c2<-1, 1> expression, after simplification I am left with <w1, w2>.

So, the expressions for c1 and c2 we found were in terms of w1 and w2, which, when multiplied by the v1 and v2 expressions, did indeed yield the final components of w1 and w2. Does this mean that it has been proven?

vela

Yup, you were asked to show you can find c₁ and c₂ for any w, and you found formulas that do exactly that.