MHB Jordan Normal Form of Unitary Transformation

[Sudharaka]

Hi everyone, :)

Recently I encountered the following problem. Hope you can confirm whether my method is correct. My answer seems so trivial and I have doubts whether it is correct.

Problem:

Find the Jordan normal form of a unitary linear transformation.

My Solution:

Now if we take the matrix of a unitary linear transformation (say \(A\)) it could be diagonalized; \(A=VDV^*\), where \(D\) is a diagonal unitary matrix and \(V\) is a unitary matrix. So therefore the Jordan normal form is obviously \(D\), where the diagonal elements consist of the eigenvalues of \(A\).

 

[Opalg]

Hi everyone, :)

Recently I encountered the following problem. Hope you can confirm whether my method is correct. My answer seems so trivial and I have doubts whether it is correct.

Problem:

Find the Jordan normal form of a unitary linear transformation.

My Solution:

Now if we take the matrix of a unitary linear transformation (say \(A\)) it could be diagonalized; \(A=VDV^*\), where \(D\) is a diagonal unitary matrix and \(V\) is a unitary matrix. So therefore the Jordan normal form is obviously \(D\), where the diagonal elements consist of the eigenvalues of \(A\).

Yes, that is absolutely correct. The result is not trivial, but all the difficulty is hidden in the statement that a unitary matrix can be diagonalised. If you are allowed to quote that, then the rest is easy. (You could add that the diagonal elements of the Jordan form must all have absolute value $1$.)

 

[Sudharaka]

Yes, that is absolutely correct. The result is not trivial, but all the difficulty is hidden in the statement that a unitary matrix can be diagonalised. If you are allowed to quote that, then the rest is easy. (You could add that the diagonal elements of the Jordan form must all have absolute value $1$.)

Thanks very much for the detailed explanation. I really appreciate it. :)