Linear Algebra - Normal and Unitary Matrices

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SUMMARY

The discussion centers on proving that a normal transformation T in a finite-dimensional unitary space V, where T-1 = -T, is a unitary transformation. It is established that T is unitary if it is normal and the absolute value of its eigenvalues is 1. The proof confirms that T2 = -I leads to eigenvalues a = ±1, thus satisfying the conditions for T to be unitary. The definition of a unitary transformation is clarified as T · T* = T* · T = I.

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Homework Statement


Suppose V is a unitary space [over C] and T: V -> V is a normal transformation that satisfies T-1=-T. Prove that T is unitary transformation.

Homework Equations


I know that T is unitary if and only if it is normal and the absolute value of its eigenvalues is 1. [*2]

The Attempt at a Solution


T-1=-T so T2=-I, now suppose a is an eigenvalue of T so T2v=a2v=-Iv what in turn means a=\pm 1.
So from [*2] we can conclude that T is unitary.

Have I missed something?
 
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Did you prove normality? Also, can you be more specific about V? Is it just a vector space? A Banach Space? Hilbert Space? Is it finite or infinite dimension? What definition of unitary are you using?
 
Thanks for posting!

Normality is given.
V is a general unitary space. [it means V is a general space over the complex field]
Forgot to mention: V is finite dimension space.
Unitary definition:T \cdot T^*=T^* \cdot T=I
 
Ah yes, I see now that it was given. It looks fine to me, though I have to say I've never heard of a "general unitary space." Perhaps this is okay in this instance since \mathbb C is algebraically self-dual, but normally one requires at the very least a normed vector space.
 
So I guess the prove is ok? [This question was formulated by myself so it is totally possible that I didn't define the question itself well]

What is normed vector space?
 
It looks fine so long as the result about being normal with modulus 1 is true. I think it's true, but haven't run through the proof myself. In any case, I would say you're good to go.

A normed vector space is a vector space with a norm. :) Not sure I can be more explicit than that.
 
Kreizhn said:
It looks fine so long as the result about being normal with modulus 1 is true. I think it's true, but haven't run through the proof myself. In any case, I would say you're good to go.

A normed vector space is a vector space with a norm. :) Not sure I can be more explicit than that.

Thanks!

Kreizhn said:
...
A normed vector space is a vector space with a norm. :) Not sure I can be more explicit than that.
I though every vector space has a norm, so I should not explicitly say it.
 

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