# How to start this Vector Space Property Proof?

1. Oct 10, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Let |v> ∈ V with |v> ≠ |0>, and let λ, μ ∈ ℂ.
Prove that if λ|v> = μ|v>, then λ = μ

2. Relevant equations
Vector Space Axioms

3. The attempt at a solution
I am struggling to begin with this one.
I can think of tons of different ways to begin, but all seem to get into a hazy area where I am unsure.
I could start by adding or "subtracting" from both sides λ|v> or by adding 0 to both sides, or any other traditional proof ways.

Is there any hint to help me go in the right direction?

2. Oct 10, 2015

### Ray Vickson

PF rules: show us your attempt first, then we are allowed to give hints.

3. Oct 10, 2015

### RJLiberator

I can give you a first attempt, I have many disaster steps.
It should be a simple proof, I am just seemingly taking the wrong first steps.

λ|v> = μ|v> (this is what we have)
First step is the challenging part: I don't know, let's try adding λ|v> to both sides.
λ|v> + λ|v> = μ|v> + λ|v>
(λ+λ)|v> = (μ+λ) |v>
Now we see that λ+λ = μ+λ
And so with simple subtraction we get λ = μ

Is that a correct way of doing it? I feel like I am skidding by something here...

4. Oct 10, 2015

### Staff: Mentor

I would suggest subtracting μ|v> from both sides (or adding -μ|v> to both sides).
This would give you λ|v> - μ|v> = |0>
(I have to admit, I'm not very familiar with this notation. It seems to be used more by physicists than by mathematicians, AFAIK.)

5. Oct 10, 2015

### RJLiberator

If I were to subtract, the proof looks as follows:
λ|v> = μ|v>
λ|v> - μ|v> = μ|v> - μ|v>
λ|v> - μ|v> = 0

But isn't this essentially what we started with?
How do we go from here to prove that λ = μ ?
Do we say:

λ|v> - μ|v> = 0
(λ-μ)|v> = 0 and since v ≠ 0 then λ-μ must = 0, and so it follows that λ = μ.

That seems like a correct path! Excellent.

It's as if my proof writing abilities are increased tenfold when I write a post on Physics Forums.

6. Oct 10, 2015

### Ray Vickson

No: if you have not yet proven that $\lambda |v \rangle = \mu |v \rangle \Longrightarrow \lambda = \mu$, then you have no reason to suppose that
$(\lambda + \lambda) |v \rangle = (\lambda + \mu) |v \rangle \Longrightarrow \lambda + \lambda = \lambda + \mu$. Basically, you are just assuming the result you want to prove.

7. Oct 10, 2015

### RJLiberator

If we have (λ-μ) |v> = 0
and since |v> ≠0, we can say (λ-μ) must equal 0, correct?

Or is this still assuming the result?

8. Oct 10, 2015

### Staff: Mentor

Correct. You that λ|v> - μ|v> = (λ-μ) |v> because of one of the vector space axioms that pertain to scalar multiplication.
Assuming the result would be assuming that λ = μ.

9. Oct 10, 2015

### Ray Vickson

No, probably not, provided you have already proven that $c |v\rangle = |0 \rangle$ implies $c = 0$, or else have assumed it as an axiom.

10. Oct 10, 2015

### RJLiberator

Ah, I see the difference.

In my first proof we had λ+λ = λ+μ
Where I was assuming the result.

In this last proof we have λ-μ = 0
Where we are not assuming anything here, we are showing what is needed.