1. Length of an eigenvector

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

2. Originally Posted by ironz
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
Length 1 means it is a unit vector. An eigen vector is by definition a non-zero vector, but of otherwise arbitary magnitude, asking for one of "length 1" is asking for a specific normalisation.

CB

3. Originally Posted by CaptainBlack
Length 1 means it is a unit vector. An eigen vector is by definition a non-zero vector, but of otherwise arbitary magnitude, asking for one of "length 1" is asking for a specific normalisation.

CB
Sorry but what does "specific normalisation" mean?

If they say "with length 1", does that mean if I find a zero vector, I have done something wrong, or I do not include that in my answer?

4. Originally Posted by ironz
Sorry but what does "specific normalisation" mean?

If they say "with length 1", does that mean if I find a zero vector, I have done something wrong, or I do not include that in my answer?
Let's assume that you are talking about eigenvalues/eigenvectors of matrices. Note that if $A$ and $\lambda$ is an eigenvalue of $A$ with associated eigenvector $x\in\mathbb{R}^n$ (or any normed vector space) then $\displaystyle A\frac{x}{\|x\|}=\frac{1}{\|x\|}Ax=\frac{\lambda x}{\|x\|}=\lambda\left(\frac{x}{\|x\|}\right)$ so that $\displaystyle \frac{x}{\|x\|}$ is also an eigenvector of $A$ associated with $\lambda$ and a quick check shows that $\displaystyle \left\|\frac{x}{\|x\|}\right\|=1$. Make sense?\

Remark: It was important here to note that $x\ne\bold{0}$ by the definition of eigenvector.

5. Originally Posted by Drexel28
Let's assume that you are talking about eigenvalues/eigenvectors of matrices. Note that if $A$ and $\lambda$ is an eigenvalue of $A$ with associated eigenvector $x\in\mathbb{R}^n$ (or any normed vector space) then $\displaystyle A\frac{x}{\|x\|}=\frac{1}{\|x\|}Ax=\frac{\lambda x}{\|x\|}=\lambda\left(\frac{x}{\|x\|}\right)$ so that $\displaystyle \frac{x}{\|x\|}$ is also an eigenvector of $A$ associated with $\lambda$ and a quick check shows that $\displaystyle \left\|\frac{x}{\|x\|}\right\|=1$. Make sense?\

Remark: It was important here to note that $x\ne\bold{0}$ 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.

6. Originally Posted by ironz
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 $\mathbb{R}^n$, namely $\displaystyle \|(x_1,\cdots,x_n)\|=\sqrt{\sum_{k=1}^{n}x_k^2}$...and something is of "length one" if it's norm is one. So, for example if $x$ is a point on the unit circle we know it's of the form $\left(\sin(\theta),\cos(\theta)\right)$ and so by definition $\|x\|=\sqrt{\sin^2(\theta)+\cos^2(\theta)}=\sqrt{1 }=1$ so that $x$ has length one. So, anyways, my point was if you find ANY eigenvector $x$ associated with eigenvalue $\lambda$ then $\alpha x$ is an eigenvector associated with $\lambda$ for any $\alpha\ne 0$. So, if $x$ is an eigenvalue associated with $\lambda$ then so is $\frac{1}{\|x\|}x$ but it's easy to show that $\displaystyle \left\|\frac{1}{\|x\|}x\right\|=1$ so that $\frac{1}{\|x\|}x$ is an eigenvalue associated with $\lambda$ of length one.I was just trying to give you an easy way to find examples.

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unit eigenvector

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