How to diagonalize a matrix with complex eigenvalues?

[Haorong Wu]

Homework Statement

Diagonalize the matrix $$ \mathbf {M} =
\begin{pmatrix}
1 & -\varphi /N\\
\varphi /N & 1\\
\end{pmatrix}
$$ to obtain the matrix $$ \mathbf{M^{'}= SMS^{-1} }$$

Homework Equations

First find the eigenvalues and eigenvectors of $\mathbf{M}$, and then normalize the eigenvectors to get $\mathbf{S^{-1}}$.

The Attempt at a Solution

After calculation, the eigenvalues are $\lambda = 1 \pm \frac \varphi N$, and the corresponding eigenvectors (unnormalized) are $\begin{pmatrix} i \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -i \\1 \end{pmatrix}$.

Then I try to Schmidt them, but I failed to normalize them.

Could you help me normalize them?

Regards

 

[member 587159]

Normalize = divide by their norm to get a vector with length 1.

The norm of the vector $(i,1)$ is $\sqrt{2}$, so we get the normalized vector $(i/\sqrt{2}, 1/\sqrt{2})$. Can you do the same thing for the other vector now?

 

[Haorong Wu]

Normalize = divide by their norm to get a vector with length 1.

The norm of the vector $(i,1)$ is $\sqrt{2}$, so we get the normalized vector $(i/\sqrt{2}, 1/\sqrt{2})$. Can you do the same thing for the other vector now?

Thanks, Math_QED.

I concluded the normalized vectors are $\frac {\left( i , 1\right) } {\sqrt 2}$ and $\frac {\left( -i , 1\right) } {\sqrt 2}$, as well. Then $\mathbf {S^{-1}} = \frac 1 {\sqrt 2} \begin{pmatrix} i & -i \\ 1 & 1 \end{pmatrix}$.
However, $\mathbf {SMS^{-1}} =\frac 1 {\sqrt 2} \begin{pmatrix} i & -i \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -\varphi /N\\ \varphi /N & 1\\ \end{pmatrix} \frac 1 {\sqrt 2} \begin{pmatrix} -i & 1 \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & -i\varphi /N \\-\varphi /N & 1 \end{pmatrix}$ , which is not the diagonized matrix I should get.
There is definitely somewhere I am wrong, I can't find it out.
Could you please help me point it out?
Thanks!

 

[Dick]

Thanks, Math_QED.

I concluded the normalized vectors are $\frac {\left( i , 1\right) } {\sqrt 2}$ and $\frac {\left( -i , 1\right) } {\sqrt 2}$, as well. Then $\mathbf {S^{-1}} = \frac 1 {\sqrt 2} \begin{pmatrix} i & -i \\ 1 & 1 \end{pmatrix}$.
However, $\mathbf {SMS^{-1}} =\frac 1 {\sqrt 2} \begin{pmatrix} i & -i \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -\varphi /N\\ \varphi /N & 1\\ \end{pmatrix} \frac 1 {\sqrt 2} \begin{pmatrix} -i & 1 \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & -i\varphi /N \\-\varphi /N & 1 \end{pmatrix}$ , which is not the diagonized matrix I should get.
There is definitely somewhere I am wrong, I can't find it out.
Could you please help me point it out?
Thanks!

You computed $S^{-1}MS$ not $SMS^{-1}$, which is the one you want.

 

[Haorong Wu]

You computed $S^{-1}MS$ not $SMS^{-1}$, which is the one you want.

Oh, no! How embarrassing! I waste all the afternoon.

Thank you, Dick, and Math_QED.

 

[Ray Vickson]

Homework Statement

Diagonalize the matrix $$ \mathbf {M} =
\begin{pmatrix}
1 & -\varphi /N\\
\varphi /N & 1\\
\end{pmatrix}
$$ to obtain the matrix $$ \mathbf{M^{'}= SMS^{-1} }$$

Homework Equations

First find the eigenvalues and eigenvectors of $\mathbf{M}$, and then normalize the eigenvectors to get $\mathbf{S^{-1}}$.

The Attempt at a Solution

After calculation, the eigenvalues are $\lambda = 1 \pm \frac \varphi N$, and the corresponding eigenvectors (unnormalized) are $\begin{pmatrix} i \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -i \\1 \end{pmatrix}$.

Then I try to Schmidt them, but I failed to normalize them.

Could you help me normalize them?

Regards

The eigenvalues of A are $1 \pm i \varphi/N,$ not what you wrote. Your typo confused me, as I wondered how a real matrix with real eigenvalues could need complex eigenvectors.