Find a unitary matrix U such that U*AU is diagonal

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The discussion focuses on finding a unitary matrix U such that U*AU is diagonal for the given matrix A. The eigenvalues identified are -1, -1, -1, and 7, with corresponding eigenvectors normalized to form the matrix U. The initial arrangement of eigenvectors in U does not yield a diagonal matrix when multiplied by A, prompting the need to reorder the eigenvectors according to their eigenvalues. The correct configuration of U is essential for achieving the diagonalization of matrix A.

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For the following matrix A, find a unitary matrix U such that U*AU is diagonal:
A =
1 2 2 2
2 1 2 2
2 2 1 2
2 2 2 1

I found the eigenvalues to be -1,-1,-1,7
and the eigenvectors to be (v1)=(-1,1,0,0),(v2)=(-1,0,1,0),(v3)=(-1,0,0,1),(v4)=(1,1,1,1)
Normalize these vectors: ||(v1)||=sqrt(2),||(v2)||=sqrt(2), ||(v3)||=sqrt(2), ||(v4)||=2
So a unitary matrix is
U=
1/sqrt(2) -1/sqrt(2) -1/sqrt(2) -1/sqrt(2)
1/sqrt(2) 0 0 1/2
0 1/sqrt(2) 0 1/2
0 0 1/sqrt(2) 1/2

But this does not satisfy U*AU is diagonal, so I'm thinking I want to change the order of the vectors. But how do I know which one is satisfies the condition? (trial and error is rather tedious)
 
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U is simply the columns of eigenvectors arranged one next to the other, in the same order as the eigenvalues for the diagonal matrix.

For example, if the diagonal matrix:
D=U^{-1}AU= <br /> \displaystyle\left[ {\begin{array}{*{20}{c}} <br /> -1&amp;0&amp;0&amp;0 \\ <br /> 0&amp;-1&amp;0&amp;0 \\ <br /> 0&amp;0&amp;-1&amp;0 \\ <br /> 0&amp;0&amp;0&amp;7 <br /> \end{array}} \right]
Then, the corresponding matrix U will have the first column as the eigenvector corresponding the eigenvalue, \lambda_1 = -1, and so on, with the last column of matrix U as the eigenvector corresponding to \lambda_4 = 7.
 
Last edited:

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