Can a Non-Diagonal Hermitian Matrix be Diagonalized Using Unitary Matrix?

[LagrangeEuler]

Every hermitian matrix is unitary diagonalizable. My question is it possible in some particular case to take hermitian matrix ##A## that is not diagonal and diagonalize it
[tex]UAU=D[/tex]
but if ##U## is not matrix that consists of eigenvectors of matrix ##A##. ##D## is diagonal matrix.

 

[anuttarasammyak]

[tex]U=U^{-1}[/tex]
I am not sure of such a specific case.

 

[LagrangeEuler]

Yes here I am talking about case ##U=U^{-1}##. I am also not sure. But for me it is interesting.

 

[jedishrfu]

is the identity matrix a trivial case?

 

[anuttarasammyak]

As an example in 2X2 marix
[tex]U=
\begin{pmatrix}
k & \alpha \\
\beta & -k \\
\end{pmatrix}
[/tex]
where
[tex]\alpha \beta = 1-k^2[/tex]
would produce non diagonal matrix A from diagonal matrix D which has two different eigenvalues with A=UDU.

 

[LagrangeEuler]

You can give me 2x2 example. But specify ##A##, ##U## and ##D##. Because ##U## still possibly can be formed of eigenvectors of ##A## in your example of ##U##.

 

[anuttarasammyak]

Eingenvalues of U are ##\pm 1##. You can pick up the cases that eigenvalues of D and A are not 1 nor -1 for your purpose.

 

[LagrangeEuler]

is the identity m

No, because the identity matrix is diagonal.

 

[LagrangeEuler]

Eingenvalues of U are ##\pm 1##. You can pick up the cases that eigenvalues of D and A are not 1 nor -1 for your purpose.

Why? Eigenvalue can be for instance ##i##.

 

[anuttarasammyak]

My #7 talks abot U of post #5. There for U [tex]\lambda^2 =1[/tex]
[tex]UDU=
\begin{pmatrix}
k & \alpha \\
\beta & -k \\
\end{pmatrix}
\begin{pmatrix}
d_1 & 0 \\
0 & d_2 \\
\end{pmatrix}
\begin{pmatrix}
k & \alpha \\
\beta & -k \\
\end{pmatrix}
=
\begin{pmatrix}
d_2+(d_1-d_2)k^2 & k\alpha (d_1-d_2) \\
k \beta (d_1-d_2) & d_1-(d_1-d_2)k^2 \\
\end{pmatrix}
=A[/tex]
, and
[tex]d_1,d_2 \neq -1,1[/tex]
to satisfy your further condition.