# Unitary Operator

1. Mar 31, 2012

### Shackleford

Here's the definition.
Let T be a linear operator on a finite-dimensional inner product space V. If $\|\vec{T(x)}\| = \|\vec{x}\| \\$ for all x in V, we call T a unitary operator.
The question is asking about for all x in some orthornormal basis for V. Isn't that the same as for all x in V?

2. Mar 31, 2012

### Dick

Not at all. It says "some orthonormal basis for V". It doesn't say "for all orthonormal bases for V". Think counterexample.

3. Apr 1, 2012

### Shackleford

Doesn't a basis generate all of V, though?

4. Apr 1, 2012

### Dick

Yes, a basis generates V. But ||U(a)||=||a|| and ||U(b)||=||b|| doesn't imply that ||U(a+b)||=||a+b||.

5. Apr 1, 2012

### Shackleford

I think the question is confusing me. I interpret "for all x in some orthornormal basis for V" as meaning for every linear combination x of the basis.

What kind of counterexample am I looking for? Just some orthonormal basis that's not unitary?

6. Apr 1, 2012

### Dick

Pick ANY orthonormal basis. Call it {e_1,e_2,...,e_n}. You want to define U somehow. The only condition that U has to satisfy is that ||U(e_i)||=1 for 1<=i<=n.

Last edited: Apr 1, 2012
7. Apr 1, 2012

### Shackleford

How about I take the standard basis for R2 and define

T(x1, x2) = (x1 - x2, 0).

T(1,0) = (1,0); ||T(e1)||= 1
T(0,1) = (-1,0); ||T(e2)||= 1

T(1,1) = (0,0); ||T(e1 + e2)||= 0

8. Apr 1, 2012

### Dick

Looks good to me!

9. Apr 1, 2012

### Shackleford

Thanks!

I have another question. I didn't understand property #5. Why does it sum over every entry, not just the diagonal?

http://planetmath.org/encyclopedia/TraceOfAMatrix.html [Broken]

Last edited by a moderator: May 5, 2017
10. Apr 1, 2012

### Dick

Use sigma notation to write out the product AA*, then take the trace. It just works out that way.

Last edited by a moderator: May 5, 2017