Unitary Operator: Proof & Counterexample

[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.

Let U be a linear operator on a finite-dimensional inner product space V. If $\|\vec{U(x)}\| = \|\vec{x}\| \\$ for all x in some orthonormal basis for V, must U be unitary? Justify your answer with a proof or a counterexample.

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?

 

[Dick]

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

 

[Shackleford]

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

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

 

[Dick]

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

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

 

[Shackleford]

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

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?

 

[Dick]

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?

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.

 

[Shackleford]

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.

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

 

[Dick]

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

Looks good to me!

 

[Shackleford]

Looks good to me!

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

 

[Dick]

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

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