# State vectors, projection matrices

1. Oct 8, 2007

### cscott

1. The problem statement, all variables and given/known data

How do I prove that if,

$$|\vec{u_1}><\vec{u_1}| + |\vec{u_2}><\vec{u_2}| = I$$,

where 'I' is the indentity matrix, that $u_1$ and $u_2$ are orthogonal and normalized?

Can anybody get me started?

Last edited: Oct 8, 2007
2. Oct 8, 2007

### Hurkyl

Staff Emeritus
You can't, with the information you've stated.

3. Oct 8, 2007

### cscott

Here's the exact question from the problem set, just in case:

Show that given two arbitrary vectors u1 , u2 ∈ C^2 such that the associated projectors satisfy Pu1 + Pu2 = 1 then it holds that the vectors u1, u2 are orthogonal and normalized.

Is it really impossible?

4. Oct 8, 2007

### Hurkyl

Staff Emeritus
Oh bah. I read the problem backwards.

Well, since yo'ure given an equation with I in it, it will probably be useful to throw I into every relevant equation you can think of.

5. Oct 8, 2007

### cscott

That's the thing, I can't think of any relevant equations besides the two inner product equations for a normalized vector and two orthogonal vectors.

6. Oct 8, 2007

### cscott

What if I multiplied by |u2> on the right-hand side of each term. Would I be correct in saying that in order for this new equation to hold true u1 and u2 have to be orthogonal and u2 would have to be normalized? Then I can do it again with u1.

7. Oct 8, 2007

### Hurkyl

Staff Emeritus
Well, then throw an I in them! What do you get?

8. Oct 8, 2007

### cscott

I think I would get:

<u1|u1>I = I
<u2|u2>I = I
<u1|u2>I = 0
<u2|u1>I = 0

Does this make sense? Is it incorrect to do it like I said in post #6?

If I substituted the expression in my OP for 'I' in these equations, and they still hold true does it prove what I need?

9. Oct 8, 2007

### Hurkyl

Staff Emeritus
You put the I in the wrong place. It's an operator; it should operate on things!

I suppose that you can put an I there, intending there to be scalar multiplication, but that doesn't do anything useful here.

10. Oct 8, 2007

### cscott

Ooh

<u1|I|u1> = I
<u2|I|u2> = I
<u1|I|u2> = 0
<u2|I|u1> = 0

Like that then?

11. Oct 8, 2007

### Hurkyl

Staff Emeritus
Yes -- those are the statements you want to prove!

Incidentally, this is a trick you want to take to heart; it's a very useful thing. (And not just in linear algebra; in analysis, you might use a partition of unity in similar ways)

12. Oct 8, 2007

### cscott

I'll try to work out the algebra.

...I think I'm slllllowly getting my head around this quantum stuff.

13. Oct 8, 2007

### Hurkyl

Staff Emeritus
Hrm...

This is surely what you want to do, but finishing off the proof is less straightforward than I thought it was.

14. Oct 8, 2007

### Hurkyl

Staff Emeritus
Shame on me for not fully thinking this through! You can do this... if you can first prove that |u1> and |u2> are linearly independent. (do you see why?)

15. Oct 8, 2007

### cscott

I see why |u1> and |u2> would have to be linearly independent. I'm not sure how I'd prove that, though. Wouldn't I have to do it based solely on the fact that adding the projection matrices gives the identity matrix?

16. Oct 8, 2007

### cscott

When I substitute $$|\vec{u_1}><\vec{u_1}| + |\vec{u_2}><\vec{u_2}| = I$$

into the LHS of
<u1|I|u1> = I
<u1|I|u2> = 0
I get
1 = I
<u1|u2> + <u1|u2> = 0
respectively

This doesn't seem right... ? It does imply that<u1|u2> = 0, though.

17. Oct 9, 2007

### Hurkyl

Staff Emeritus
Assume they are dependent, see what happens. Rank might be involved.

I don't see how you got that. And remember that <u1|I|u2> = 0 is what you're trying to prove.