Unitary matrices

[Zareon]

I've read somewhere that a unitary matrix U can be defined by the property:

(1) U*=U^{-1} (* = hermitian conjugate)

or by the fact that it preserves lengths of vectors:

(2) <Ux,Ux>=<x,x>

I have trouble seeing why they are equivalent.

 

It's obvious to see that (1)=> (2):

<Ux,Ux>=(Ux)*(Ux)=x*(U*U)x=x*x=<x,x>

 

But not the other way around. I CAN prove it for real vector spaces, where U is an orthogonal matrix from the fact that <v,w>=<w,v>. Then I would do:

 

<v+w,v+w>=<U(v+w),U(v+w)>=<Uv,Uv>+<Uw,Uw>+2<Uv,Uw>=<v,v>+<w,w>+2<Uv,U,w> and working out the left side gives <Uv,Uw>=<v,w>.

and from this that the columns of U are orthonormal, since [math]<Ue_i,Ue_j>=<e_i,e_j>=\delta_{ij}[/math]

 

But for a complex vector space where <v,w>=<w,v>* all the above gives is:

Re(<Uv,Uw>)=Re(<v,w>).

 

EDIT: made some mistakes

[the tree]

(1)[tex]U^\dagger=U^{-1}[/tex]

On this forum, use [math], rather than [tex], so [math]U^{\dagger}=U^{-1}[/math], or better, use the ^{and _{tags when all you need is the the sub/superscript features so: U†=U-1.}}

[Zareon]

Thanks the tree.

 

Anyway, I've found the answer. I just had to evaluate <U(v+iw),U(v+iw)> to find that also Im(<Uv,Uw>)=Im(<v,w>).

[the tree]

No problem. I wish I could have responded to your question but to be honest I don't know anything about matrices, so I figured I'd just give you some general info.