# Eigenvectors/Eigenvalues help!

• Nov 10th 2008, 08:45 AM
farso
Eigenvectors/Eigenvalues help!
Hi everyone

I have a piece of maths coursework of an engineering course which the final part of which is providing some difficulties, so was wondering if someone here would be kind enough to help please.

I am aware of what eigenvalues/vectors are, but i am not certain how to apply the things i know to this question - nor does anyone on my course or any other doctor i have asked!

Anyway, the question (or the very first part!).

Show that:

$\displaystyle \left[ \begin{array}{cccc} cos(x) & -sin(x) & 0 \\ sin(x)& cos(x) & 0 \\ 0 & 0 & 1 \end{array} \right]$

Has the an eigenvector

$\displaystyle \left[ \begin{array}{cccc} 1 \\ i \\ 0 \end{array} \right]$

Write the answer in exponential form.

Id appreciate any help on the matter

• Nov 10th 2008, 10:23 AM
Opalg
" Write the answer in exponential form" means "Use the fact that $\displaystyle \cos x+i\sin x = e^{ix}$".

Multiply the matrix by the given eigenvector to get $\displaystyle \begin{bmatrix} \cos x& -\sin x& 0 \\ \sin x& \cos x& 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1\\i\\0\end{bmatrix}= \begin{bmatrix}\cos x-i\sin x\\ \sin x + i\cos x \\0\end{bmatrix}= \begin{bmatrix}e^{-ix}\\ ie^{-ix} \\0\end{bmatrix} = e^{-ix}\begin{bmatrix}1\\i\\0\end{bmatrix}$.

Now you should be able to see what the eigenvalue is.
• Nov 10th 2008, 11:16 AM
farso
Hi

Thanks for the fast reply. I understand the first part of your reply, but i really dont understand how to get an eigenvalue from this.

Sorry for appearing to not know a thing here really, I am really keen to understand all this though as its essential for my engineering degree.

Thanks
• Nov 10th 2008, 12:52 PM
Opalg
Quote:

Originally Posted by farso
I really don't understand how to get an eigenvalue from this.

The definition of an eigenvalue of a matrix A is that it is a number $\displaystyle \lambda$ such that for some nonzero vector v, the equation $\displaystyle Av = \lambda v$ holds.

In this problem, if we call the matrix A, and v is the vector $\displaystyle \mathbf{v}=\begin{bmatrix}1\\i\\0\end{bmatrix}$, then the calculation in my previous comment shows that $\displaystyle A\mathbf{v} = e^{-ix}\mathbf{v}$. That shows that v is an eigenvector for the eigenvalue $\displaystyle \lambda = e^{ix}$.
• Nov 13th 2008, 01:57 AM
farso
Thanks very much, it tok me a fair bit of work to get my head around $\displaystyle iSinx + Cos x$ in exponential form, but finally got it.
Perhaps you can help me with another part of the same question?

If the first matrix i said was $\displaystyle Rz$
and
$\displaystyle Ry = \left[ \begin{array}{cccc} cos(x) & 0 & sin x \\ 0 & 1 & 0 \\ -sin x & 0 & cos x \end{array} \right]$
which are both used to represent movment of a robot arm starting at.
$\displaystyle \left[ \begin{array}{cccc} 1 \\ 0 \\ 0 \end{array} \right]$
and its movment is applied pi/2 to z axis followed by the y axis
show that it ends in position
$\displaystyle \left[ \begin{array}{cccc} 0 \\ 1 \\ 0 \end{array} \right]$

• Nov 16th 2008, 07:01 AM
geo3
Quote:

Originally Posted by Opalg

Multiply the matrix by the given eigenvector to get $\displaystyle \begin{bmatrix} \cos x& -\sin x& 0 \\ \sin x& \cos x& 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1\\i\\0\end{bmatrix}= \begin{bmatrix}\cos x-i\sin x\\ \sin x + i\cos x \\0\end{bmatrix}= \begin{bmatrix}e^{-ix}\\ ie^{-ix} \\0\end{bmatrix} = e^{-ix}\begin{bmatrix}1\\i\\0\end{bmatrix}$.

doesn't sinx + icosx = ie^ix ???
• Nov 16th 2008, 07:06 AM
Opalg
Quote:

Originally Posted by geo3
doesn't sinx + icosx = ie^ix ???

No, $\displaystyle ie^{ix} = i(\cos x+i\sin x) = i\cos x - \sin x$.