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
UAU=D
but if $U$ is not matrix that consists of eigenvectors of matrix $A$. $D$ is diagonal matrix.

 

[anuttarasammyak]

U=U^{-1}
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
U=<br /> \begin{pmatrix}<br /> k &amp; \alpha \\<br /> \beta &amp; -k \\<br /> \end{pmatrix}<br />
where
\alpha \beta = 1-k^2
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 \lambda^2 =1
UDU=<br /> \begin{pmatrix}<br /> k &amp; \alpha \\<br /> \beta &amp; -k \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> d_1 &amp; 0 \\<br /> 0 &amp; d_2 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> k &amp; \alpha \\<br /> \beta &amp; -k \\<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> d_2+(d_1-d_2)k^2 &amp; k\alpha (d_1-d_2) \\<br /> k \beta (d_1-d_2) &amp; d_1-(d_1-d_2)k^2 \\<br /> \end{pmatrix}<br /> =A
, and
d_1,d_2 \neq -1,1
to satisfy your further condition.