Thread: Euclidean norm of U*x when norm of x is 1!

Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

||y|| = 1 if and only if ||x|| = 1.

I can't prove this statement and don't know where to start!

Thanks so much!

Originally Posted by Damien

Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

||y|| = 1 if and only if ||x|| = 1.

I can't prove this statement and don't know where to start!

Thanks so much!

The unitary matrix has at least one complex eigenvalue; what do you know about the eigenvalues of a unitary matrix? U is unitary if and only if U* is unitary right? Then since U*x = ax for some eigenvalue a, take the norm on both sides and use some properties of the norm to get what you want.

Originally Posted by Damien

Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

||y|| = 1 if and only if ||x|| = 1.

I can't prove this statement and don't know where to start!

Thanks so much!

let be the columns of since is unitary, we have where is the Kronecker's delta. let be the entries of the vectors respectively.

now we have and thus and thus

Thanks a lot guys. 2 different proofs, but both very nice.

1.
Every matrix has at least one eigenvalue (fundamental theorem of algebra)
If U is unitary then U* is unitary too
Every eigenvalue of a unitary matrix has absolute value of 1
So: ||y|| = ||U*x|| = ||lambda x|| = |lambda| · ||x|| = ||x||

2.
If y = U*x then x = Uy
So: ||x||^2 = x*x = y*U*Uy = y*y = ||y||^2 <=> ||x|| = ||y||

Thanks a lot to both of you, really appreciate it!