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