I have been told that for each eigenvalue, find an eigenvector of length 1.
How would I do this? What does "length 1" mean?
Thank you
Let's assume that you are talking about eigenvalues/eigenvectors of matrices. Note that if $\displaystyle A$ and $\displaystyle \lambda$ is an eigenvalue of $\displaystyle A$ with associated eigenvector $\displaystyle x\in\mathbb{R}^n$ (or any normed vector space) then $\displaystyle \displaystyle A\frac{x}{\|x\|}=\frac{1}{\|x\|}Ax=\frac{\lambda x}{\|x\|}=\lambda\left(\frac{x}{\|x\|}\right)$ so that $\displaystyle \displaystyle \frac{x}{\|x\|}$ is also an eigenvector of $\displaystyle A$ associated with $\displaystyle \lambda$ and a quick check shows that $\displaystyle \displaystyle \left\|\frac{x}{\|x\|}\right\|=1$. Make sense?\
Remark: It was important here to note that $\displaystyle x\ne\bold{0}$ by the definition of eigenvector.
Sorry, no, I don't understand. What does the double modulus sign mean and what is x?
Also, if for the purposes of making it easy for me to type, we let X equal the double modulus sign of x, then how do you go from
(1/X)(Ax) to lambda x / X?
Finally, what does x / X mean?
Sorry if I seem a bit slow on this, we didn't have the best of lecturers and I made the mistake of not using my tutorials to their full potential.
The double modulus sign is the usual norm on $\displaystyle \mathbb{R}^n$, namely $\displaystyle \displaystyle \|(x_1,\cdots,x_n)\|=\sqrt{\sum_{k=1}^{n}x_k^2}$...and something is of "length one" if it's norm is one. So, for example if $\displaystyle x$ is a point on the unit circle we know it's of the form $\displaystyle \left(\sin(\theta),\cos(\theta)\right)$ and so by definition $\displaystyle \|x\|=\sqrt{\sin^2(\theta)+\cos^2(\theta)}=\sqrt{1 }=1$ so that $\displaystyle x$ has length one. So, anyways, my point was if you find ANY eigenvector $\displaystyle x$ associated with eigenvalue $\displaystyle \lambda$ then $\displaystyle \alpha x$ is an eigenvector associated with $\displaystyle \lambda$ for any $\displaystyle \alpha\ne 0$. So, if $\displaystyle x$ is an eigenvalue associated with $\displaystyle \lambda$ then so is $\displaystyle \frac{1}{\|x\|}x$ but it's easy to show that $\displaystyle \displaystyle \left\|\frac{1}{\|x\|}x\right\|=1$ so that $\displaystyle \frac{1}{\|x\|}x$ is an eigenvalue associated with $\displaystyle \lambda$ of length one.I was just trying to give you an easy way to find examples.