# How to diagonalize a matrix with complex eigenvalues?

## 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

Related Calculus and Beyond Homework Help News on Phys.org
Math_QED
Homework Helper
2019 Award
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?

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 definetely somewhere I am wrong, I can't find it out.
Thanks!

Dick
Homework Helper
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 definetely somewhere I am wrong, I can't find it out.
Thanks!
You computed ##S^{-1}MS## not ##SMS^{-1}##, which is the one you want.

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 Helper
Dearly Missed

## 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.