# Euclidean norm of U*x when norm of x is 1!

• Oct 1st 2009, 10:26 AM
Damien
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! (Clapping)
• Oct 1st 2009, 12:52 PM
qspeechc
Quote:

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! (Clapping)

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.
• Oct 1st 2009, 08:29 PM
NonCommAlg
Quote:

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! (Clapping)

let $e_1, \cdots , e_n$ be the columns of $U.$ since $U$ is unitary, we have $e_i \cdot \bar{e_j}=\delta_{ij},$ where $\delta_{ij}$ is the Kronecker's delta. let $x_i, y_i, \ \ 1 \leq i \leq n,$ be the entries of the vectors $x,y$ respectively.

now we have $\sum_{i=1}^n y_ie_i = Uy=x$ and thus $||x||^2=||Uy||^2=\left(\sum_{i=1}^ny_ie_i \right) \cdot \left(\sum_{j=1}^n \bar{y_j} \bar{e_j} \right)=\sum_{1 \leq i,j \leq n}y_i \bar{y_j} \delta_{ij}=\sum_{i=1}^n |y_i|^2=||y||^2$ and thus $||x||=||y||.$
• Oct 2nd 2009, 12:45 AM
Damien
Thank you!
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!