MHB Eigenvectors of 2*2 rotation matrix

[bugatti79]

Hi Folks,

I calculate the eigenvalues of \begin{bmatrix}\cos \theta& \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} to be \lambda_1=e^{i \theta} and \lambda_2=e^{-i \theta}

for \lambda_1=e^{i \theta}=\cos \theta + i \sin \theta I calculate the eigenvector via A \lambda = \lambda V as

\begin{bmatrix}\cos -(\cos \theta+ i \sin \theta) & \sin \theta \\ - \sin \theta & \cos -(\cos \theta+ i \sin \theta)\end{bmatrix} \begin{bmatrix}v_1\\ v_2\end{bmatrix}=\vec{0}

which reduces to

- i \sin \theta v_1+ \sin \theta v_2=0
-\sin \theta v_1-i \sin \theta v_2=0

I am stumped at this point...how shall I proceed?

 

[Opalg]

Hi Folks,

I calculate the eigenvalues of \begin{bmatrix}\cos \theta& \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} to be \lambda_1=e^{i \theta} and \lambda_2=e^{-i \theta}

for \lambda_1=e^{i \theta}=\cos \theta + i \sin \theta I calculate the eigenvector via A \lambda = \lambda V as

\begin{bmatrix}\cos -(\cos \theta+ i \sin \theta) & \sin \theta \\ - \sin \theta & \cos -(\cos \theta+ i \sin \theta)\end{bmatrix} \begin{bmatrix}v_1\\ v_2\end{bmatrix}=\vec{0}

which reduces to

- i \sin \theta v_1+ \sin \theta v_2=0
-\sin \theta v_1-i \sin \theta v_2=0

I am stumped at this point...how shall I proceed?

Divide those equations by $\sin\theta$ (assuming that $\sin\theta\ne0$).

 

[bugatti79]

Divide those equations by $\sin\theta$ (assuming that $\sin\theta\ne0$).

Then we get

- i v_1+ v_2=0 (1)
- v_1-i v_2=0 (2)

v_2=i v_1 from 1

v_2=-v_1/i from 2

1) These contradict? How is the eigenvector obtained from this?

2) what if we have a situation where \theta=0? Then \sin \theta=0

 

[Opalg]

Then we get

- i v_1+ v_2=0 (1)
- v_1-i v_2=0 (2)

v_2=i v_1 from 1

v_2=-v_1/i from 2

1) These contradict? How is the eigenvector obtained from this?

2) what if we have a situation where \theta=0? Then \sin \theta=0

1) They don't contradict, because $i^2=-1$ and so $-1/i = i$.

2) If $\theta=0$ then the matrix becomes $\begin{bmatrix}1&0 \\0&1 \end{bmatrix}$ (the identity matrix), with a repeated eigenvalue $1$.

 

[bugatti79]

1) They don't contradict, because $i^2=-1$ and so $-1/i = i$.

2) If $\theta=0$ then the matrix becomes $\begin{bmatrix}1&0 \\0&1 \end{bmatrix}$ (the identity matrix), with a repeated eigenvalue $1$.

1) Ok, so the eigenvector for

\lambda_1=e^{i \theta} is \begin{bmatrix}1\\ i\end{bmatrix}

and

\lambda_2=e^{-i \theta} is \begin{bmatrix}1\\ 1/i\end{bmatrix}

To show these 2 vectors are orthogonal I get the inner product

<v_1,v_2>=(1*1)+(i*1/i)\ne 0 but I expect 0...?

 

[Opalg]

1) Ok, so the eigenvector for

\lambda_1=e^{i \theta} is \begin{bmatrix}1\\ i\end{bmatrix}

and

\lambda_2=e^{-i \theta} is \begin{bmatrix}1\\ 1/i\end{bmatrix}

To show these 2 vectors are orthogonal I get the inner product

<v_1,v_2>=(1*1)+(i*1/i)\ne 0 but I expect 0...?

The definition of the inner product of two complex vectors is that you have to take the complex conjugate of the second one: if $x = ( x_1,x_2)$ and $y = (y_1,y_2)$ then $\langle x,y\rangle = x_1\overline{y_1} + x_2\overline{y_2}.$