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Math Help - Eigenvector, Eigenvalue and diagonalising help

  1. #1
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    Eigenvector, Eigenvalue and diagonalising help

    Hi guys, i've got some issues at the moment regarding eigenvector and eigenvalue Your help will be greatly appreciated. Thanks =)

    First the matrix c is given as:

    3 1 0
    2 2 0
    1 6 5 (Note: i have shown that the eigenvalues are 1,4 and 5).

    It then asked me to write out the polynomial of C, namely write out: determinant(C - lamda multiply the identity matrix) as a polynomial of lamda. Any ideas whats this about?


    Lastly, there are 4 column vectors:

    a:

    3
    2
    -4

    b:

    -6
    -4
    8

    c:
    0
    -5
    0

    d:
    0
    0
    -5

    The sets {ac}, {ad}, {bc},d}, {bd}, and {cd} are all linearly independent.

    The question asked me to find a basis of the vector space R^3 which lies in {a b c d}

    any ideas guys?

    thanks. Your help is much appreciated.
    Last edited by Redeemer_Pie; March 18th 2009 at 08:20 AM. Reason: figured out diagonlisation and left out a previous inquiry
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  2. #2
    Super Member Showcase_22's Avatar
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    It wants you to first find the characteristic polynomial. This is found by doing |C-xI_3|=0.

    ie. \begin{vmatrix}<br />
{3}&{1}&{0}\\ <br />
{2}&{2}&{0}\\ <br />
{1}&{6}&{5}<br />
\end{vmatrix}-x\begin{vmatrix}<br />
{1}&{0}&{0}\\ <br />
{0}&{1}&{0}\\ <br />
{0}&{0}&{1}<br />
\end{vmatrix}=\begin{vmatrix}<br />
{3-x}&{1}&{0}\\ <br />
{2}&{2-x}&{0}\\ <br />
{1}&{6}&{5-x}<br />
\end{vmatrix}

    Normally this would be the way to do it however, you've already found the eigenvalues. The eigenvalues are the roots of the characteristic polynomial. Therefore the characteristic polynomial is p_C(\lambda)=(\lambda-1)(\lambda-4)(\lambda-5)=0.

    In the case of the four column vectors it's important to notice that since you're dealing with \mathbb{R}^3 and you have 4 vectors which span. As a result one of the vectors must be a linear combination of the others.

    Sifting the vectors gives a,c and d. b is removed because it is -2\begin{pmatrix}<br />
{3}\\ <br />
{2}\\ <br />
{-4}<br />
\end{pmatrix} or -2a (ie. a linear combination of a vector you already have as part of your basis).
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  3. #3
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    Diagonalisation help!

    Hi guys,

    i though i got it correct. Turns out i've done it wrong! If you could show me what to do that'll be great! Thanks guys.

    Firstly let matrix A be:

    3 1 0
    2 2 0
    1 6 5

    I have found the eigenvalues of 1,4 and 5 are the eigenvalues of A and the eigenvectors:

    x:
    4
    -8
    11

    y:
    1
    1
    -7

    and w:
    0
    0
    1

    are corresponding to the eigenvalues 1,4,5 respectively.

    It then asked me to find a matrix lets say, T such that T^(-1)AT is diagonal. The T^(-1) is the inverse of the matrix T. Please note that the equation T^(-1)AT stands for multiplication ie. T^(-1) multiply A multiply T.

    Does anyone know how to do this? Im really stuck on this Again, your contribution is much appreciated!

    Regards =)
    Last edited by mr fantastic; March 19th 2009 at 03:34 AM. Reason: Moved from a new thread and edited accordingly
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  4. #4
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    Hello,

    Is there something in your notes that says that T is the matrix formed by the eigenvectors ?

    Let D be the diagonal matrix associated to A.

    D=\begin{pmatrix} 1 &0&0 \\ 0&4&0 \\ 0&0&5 \end{pmatrix}

    Then there is some matrix P such that A=P^{-1}DP, where P is formed by the eigenvectors (in the same order as the eigenvalues) in columns :
    P=\begin{pmatrix} 4&1&0 \\ -8&1&0 \\ 11&-7&1 \end{pmatrix}

    Thus D=PAP^{-1}, which is a diagonal matrix.

    Now I bet you can find P^{-1}

    (actually, T=P^{-1} in your problem)
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  5. #5
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    Thank you sooooo much!!!!
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  6. #6
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    Quote Originally Posted by Showcase_22 View Post

    Sifting the vectors gives a,c and d. b is removed because it is -2\begin{pmatrix}<br />
{3}\\ <br />
{2}\\ <br />
{-4}<br />
\end{pmatrix} or -2a (ie. a linear combination of a vector you already have as part of your basis).
    So what would be the basis of the vector space of R^3 which lies in a,c and d given that d is discarded? Is there an equation i have to put in? I looked through the meaning in my textbooks but its waaay advanced and dont understand the jargon
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  7. #7
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Redeemer_Pie View Post
    So what would be the basis of the vector space of R^3 which lies in a,c and d given that d is discarded?
    Do you mean "b is discarded"?

    The vector space has a basis of a,c and d.
    ie. say the basis of \mathbb{R} is denoted as B. Then B=\{a,c,d \} where a,c and d were defined in your first post.

    I don't think you have to put in an equation. You've found three vectors that form a basis of \mathbb{R}^3, you can't really put an equation to that.

    I looked through the meaning in my textbooks but its waaay advanced and dont understand the jargon
    What meaning did you look up?
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  8. #8
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    Quote Originally Posted by Showcase_22 View Post
    Do you mean "b is discarded"?

    The vector space has a basis of a,c and d.
    ie. say the basis of \mathbb{R} is denoted as B. Then B=\{a,c,d \} where a,c and d were defined in your first post.

    I don't think you have to put in an equation. You've found three vectors that form a basis of \mathbb{R}^3, you can't really put an equation to that.



    What meaning did you look up?
    I looked up bases and dimensions in my textbook but never really understood it.

    At first i thought we have to put it into some sort of equation like in my textbook like: c1v1 + c2v2 +...cnvn = (0,0,0) where v are the vectors and c a scalar or something like that. I dont know if you encountered that equation before.

    Again. Thanks for your help
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  9. #9
    Super Member Showcase_22's Avatar
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    Suppose you have a set of vectors which span a vector space.

    For example, in \mathbb{R}^3 (1,0,0),(0,1,0) and (0,0,1) span the vector space. This means that using any combination of these you can get any vector in the vector space.
    The equation you wrote down is a means of proving that these vectors are linearly independent.

    \alpha_1(1,0,0)+\alpha_2(0,1,0)+\alpha_3(0,0,1)=(0  ,0,0).
    Just reading across the rows show \alpha_1=\alpha_2=\alpha_3=0 which shows that (1,0,0),(0,1,0) and (0,0,1) are linearly independent. If a set of vectors span a vector space and are linearly independent, then they form a basis.

    The number of vectors in a basis is the dimension of the vector space. Since we have three vectors in this case, \mathbb{R}^3 has a dimension of 3.

    The same ideas can be applied to other vector spaces in any dimension. I just used the standard basis in \mathbb{R}^3 as an example.
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  10. #10
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    ok Great

    Thanks again for the help! Much appreciated =D
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