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Math Help - Length of an eigenvector

  1. #1
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    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
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ironz View Post
    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
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    Quote Originally Posted by CaptainBlack View Post
    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?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ironz View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ironz View Post
    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.
    Last edited by CaptainBlack; January 4th 2011 at 07:27 PM. Reason: fix some latex
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