Finding rotation axis and angle, using eigenvalues

[ahmed markhoos]

http://im54.gulfup.com/ZtJZd5.png

http://im54.gulfup.com/SJgJXh.png Till now compared with other subjects that I studied by my self, linear algebra is really the toughest one

Anyway, here I found a plane using characteristic eq. which really surprised me since det(M)=1, M is the matrix I've posted.

$-2x+y+z=0$ , how is that? a rotation around plane! in which dimension that's possible!, as the book did I took a vector satisfy the plane that's $(1,1,1)$, then I found $\theta = π$

Everything is correct according to book solution. But I'm not convinced yet! a plane has a lot of vectors in many directions nevertheless we derived the axis of rotation from plane eq.?! that's really weird.

Thanks a lot guys.

 

[andrewkirk]

There is no rotation around a plane. It is reflection in a plane followed by (or preceded by, as the two operations commute) a rotation around the normal to the plane (a plane only has one normal). The axis of rotation is the straight line through the origin in the direction of that normal.

 

[ahmed markhoos]

There is no rotation around a plane. It is reflection in a plane followed by (or preceded by, as the two operations commute) a rotation around the normal to the plane (a plane only has one normal). The axis of rotation is the straight line through the origin in the direction of that normal.

But the determinant shows no reflection ?

 

[andrewkirk]

But the determinant shows no reflection ?

Then in that case you don't need to bother with a plane. Just find the axis of rotation. It will be the line through the origin in the direction of the eigenvector with an associated eigenvalue of 1.

 

[ahmed markhoos]

Then in that case you don't need to bother with a plane. Just find the axis of rotation. It will be the line through the origin in the direction of the eigenvector with an associated eigenvalue of 1.

I did what you exactly said, but as you can see from the matrix in the question: "using first row"

$-x+2y+2z=\frac{1}{3}λx$

λ=1, that gives me $-2x+y+z=0$

there is more than one vector satisfy the conditions you said "through the origin and eigenvector with an associated eigenvalue of 1"

what am I really missing?, I am supposed to have a line not a plane!

 

[andrewkirk]

What eigenvector(s) have you found that has an eigenvalue of 1?

Have they taught you how to find eigenvectors?

 

[ahmed markhoos]

What eigenvector(s) have you found that has an eigenvalue of 1?

Have they taught you how to find eigenvectors?

I'm frustrated now by your question ><.

I know how to find eigenvalues and eigenvectors for 2×2 matrix with no doubt. Maybe I'm really not getting the idea of 3×3 matrices?! The book I read didn't point out clearly how to find the eigenvectors (of 3×3), just degeneracy in 3×3 matrices by Grand-Schmidt method.

Now I'm afraid that's the answer might be NO .
"Mathematical Methods in the physical sciences, by Mary Boas"

 

[andrewkirk]

Do you know how to solve simultaneous equations then? Because once you have an eigenvalue of $\lambda$ you can find an eigenvector as follows, where M is the matrix you wrote above and $\vec{v}$ is the eigenvector:

$$M\vec{v}=\lambda \vec{v}=\lambda I\vec{v}$$
Hence
$$M'\vec{v}=0$$
where $M'=M-\lambda I$ - that is, M' is the original matrix M, with $\lambda$ subtracted from each of the diagonal elements.

So just solve that second equation and you're done.

Here of course, you have $\lambda = 1$ for the eigenvalue of interest.

 

[ahmed markhoos]

Do you know how to solve simultaneous equations then? Because once you have an eigenvalue of $\lambda$ you can find an eigenvector as follows, where M is the matrix you wrote above and $\vec{v}$ is the eigenvector:

$$M\vec{v}=\lambda \vec{v}=\lambda I\vec{v}$$
Hence
$$M'\vec{v}=0$$
where $M'=M-\lambda I$ - that is, M' is the original matrix M, with $\lambda$ subtracted from each of the diagonal elements.

So just solve that second equation and you're done.

Here of course, you have $\lambda = 1$ for the eigenvalue of interest.

Do you know how to solve simultaneous equations then? Because once you have an eigenvalue of $\lambda$ you can find an eigenvector as follows, where M is the matrix you wrote above and $\vec{v}$ is the eigenvector:

$$M\vec{v}=\lambda \vec{v}=\lambda I\vec{v}$$
Hence
$$M'\vec{v}=0$$
where $M'=M-\lambda I$ - that is, M' is the original matrix M, with $\lambda$ subtracted from each of the diagonal elements.

So just solve that second equation and you're done.

Here of course, you have $\lambda = 1$ for the eigenvalue of interest.

Thanks a lot,

I was blind from seeing that obvious thing !