Eigenvectors/Eigenvalues help!

**farso** · November 10th 2008, 08:45 AM

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:

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

Has the an eigenvector

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

Write the answer in exponential form.

Id appreciate any help on the matter

thanks in advance!

**Opalg** · November 10th 2008, 10:23 AM

" Write the answer in exponential form" means "Use the fact that $\cos x+i\sin x = e^{ix}$ ".

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

Now you should be able to see what the eigenvalue is.

**farso** · November 10th 2008, 11:16 AM

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

**Opalg** · November 10th 2008, 12:52 PM

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 $\lambda$ such that for some nonzero vector v, the equation $Av = \lambda v$ holds.

In this problem, if we call the matrix A, and v is the vector $\mathbf{v}=\begin{bmatrix}1\\i\\0\end{bmatrix}$ , then the calculation in my previous comment shows that $A\mathbf{v} = e^{-ix}\mathbf{v}$ . That shows that v is an eigenvector for the eigenvalue $\lambda = e^{ix}$ .

**farso** · November 13th 2008, 01:57 AM

Thanks very much, it tok me a fair bit of work to get my head around $iSinx + Cos x$ 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 $Rz$
and
$ 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.
$ \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
$ \left[ \begin{array}{cccc} 0 \\ 1 \\ 0 \end{array} \right] $

Thanks for your help!

**geo3** · November 16th 2008, 07:01 AM

Originally Posted by Opalg

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

doesn't sinx + icosx = ie^ix ???

**Opalg** · November 16th 2008, 07:06 AM

Originally Posted by geo3

doesn't sinx + icosx = ie^ix ???

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

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