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Math Help - Jordan Normal Form Question

  1. #1
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    Jordan Normal Form Question



    So far I've found the characteristic equation.

    [Math] (t-1)^2(t+2) [/tex]

    So the eigenvalues, are 1,1,-2

    And I know the jordan matrix is along these lines(below)

    J= \begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix}

    But I literally have NO clue on where the extra 1's should be.

    I know a singular eigenvector towards T for sure, but I'm confused on -2, as the answer is totally different in my mark scheme :S!
    so.....

    \textit{}<br />
e^1 = <br />
\left( {\begin{array}{cc}<br />
1 \\<br />
1 \\<br />
1<br />
\end{array} } \right)
    \textit{}<br />
e^{-2} = <br />
\left( {\begin{array}{cc}<br />
2 \\<br />
1 \\<br />
1<br />
\end{array} } \right)

    For the last one i'm trying...
    [Math](A-I)^2\begin{bmatrix}a&b&c\end{bmatrix}=\begin{bmatr ix}1&1&1\end{bmatrix}[/tex]


    T= \begin{bmatrix}1&2&e^1\\1&1&e^1\\1&1&e^1}\end{bmat  rix}

    I also know A^5 = TJ^5T^{-1}

    I know this is a bit scrappy but could anybody help me on the bits i'm struggling with..

    1) Finding a second eigenvector for t=1, making sure my t=-2 eigenvector is correct.

    2) Explaining how you decide how many, and where to put the 1's in a jordan matrix.

    Thank you
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Monkens View Post


    So far I've found the characteristic equation.

    [Math] (t-1)^2(t+2) [/tex]

    So the eigenvalues, are 1,1,-2

    And I know the jordan matrix is along these lines(below)

    J= \begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix}

    But I literally have NO clue on where the extra 1's should be.

    I know a singular eigenvector towards T for sure, but I'm confused on -2, as the answer is totally different in my mark scheme :S!
    so.....

    \textit{}<br />
e^1 = <br />
\left( {\begin{array}{cc}<br />
1 \\<br />
1 \\<br />
1<br />
\end{array} } \right)
    \textit{}<br />
e^{-2} = <br />
\left( {\begin{array}{cc}<br />
2 \\<br />
1 \\<br />
1<br />
\end{array} } \right)

    For the last one i'm trying...
    [Math](A-I)^2\begin{bmatrix}a&b&c\end{bmatrix}=\begin{bmatr ix}1&1&1\end{bmatrix}[/tex]


    T= \begin{bmatrix}1&2&e^1\\1&1&e^1\\1&1&e^1}\end{bmat  rix}

    I also know A^5 = TJ^5T^{-1}

    I know this is a bit scrappy but could anybody help me on the bits i'm struggling with..

    1) Finding a second eigenvector for t=1, making sure my t=-2 eigenvector is correct.

    2) Explaining how you decide how many, and where to put the 1's in a jordan matrix.

    Thank you
    Oh god, actually finding the representation for the Jordan canonical form is hellacious! One question before I help, is this for a graded assignment?
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  3. #3
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    Nope it's a mock exam question from a few years back. Got my real exam tommorow.... It's always a question in that format with different numbers but I just can't suss it with my notes or by the mark scheme... =.=
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Monkens View Post
    Nope it's a mock exam question from a few years back. Got my real exam tommorow.... It's always a question in that format with different numbers but I just can't suss it with my notes or by the mark scheme... =.=
    Ok, to start to find out how many "extra" one's there are you need to recall that the geometric multiplicity of the eigenvalue \lambda is \text{null}\left(A-\lambda I\right). So, what's \text{null}\left(A-I\right)?
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  5. #5
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    Well i'm guessing we look at the eigenvalue 1, as it's the one in the brackets, and due to the bracket being to the power of two, is the null(A-I) equal to 2, so the geometric multiplicity is 2? Sorry if i'm way off here my head is just scrambled
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Monkens View Post
    Well i'm guessing we look at the eigenvalue 1, as it's the one in the brackets, and due to the bracket being to the power of two, is the null(A-I) equal to 2, so the geometric multiplicity is 2? Sorry if i'm way off here my head is just scrambled
    I get that \text{rk}\left(A-I\right)=2 and so by the Rank-Nullity theorem we have that \text{null}\left(A-I\right)=1. It follows that J=\begin{pmatrix}0 & 0 & -2\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}. So how about finding those T's?
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  7. #7
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    If null(A-I) = 2, does that have 2 1's above the diagonal??

    And yes let's move onto the T's.

    Looking at the solutions, I get the correct answer to t = 1, with (1,1,1). However when I put in -2, I get (2,1,1), whereas in the solutions it says (1,2,2).

    For the third t, where t=1 again, I have written the method I have been trying
    (A-I)^2\begin{bmatrix}a&b&c\end{bmatrix}=\begin{bmatr  ix}1&1&1\end{bmatrix}

    Is this correct I do it but still get a different answer to the solutions again..
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  8. #8
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    Since you have determined that 1 is a double root of the characteristic equation (has "algbraic multiplicity" 2) you know that the "Jordan Normal Form" is either
    \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}
    (if eigenvalue 1 has "geometric multiplicity 2) (two independent eigenvectors corresponding to eigenalue 1)
    or
    \begin{bmatrix}1  & 1 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}
    (if eigenvalue 1 has "geometric multiplicity" 1) (only one independent eigenvectors corresponding to eigenvalue 1.
    The "exra ones" only go above the multiple eigenvalue so there is only one place it could be.

    The point is this: every matrix satisfies its own characteristic equation so for this matix, A, it must b true that (A- I)^2(A- 2I)v= 0 where v is any vector. Obviously if (A- 2I)v= 0 (which is the same as saying that Av= 2v) this is true. Also if (A- I)v= 0 so that Av= v, it is true. That, is if v is an eigenvector of A with eigenvalue 2 or 1, that is true. If A has two independent eigenvectors with eigenvalue 1 (1 has geometric multiplicity 2) then, since an eigenvector corresponding to eigenvalue 2 would have to be independent of them, there are 3independent eigenvectors and the matrix T having those eigevectors as columns makes T^{-1}AT diagonal.

    If there are NOT two independent eigenvectors for eigenvalue 1 (it has geometric multiplicity 1) it still must be true that (A- I)^2(A- 2I)v= 0 for every vector. What that means is there must be an vector, independent of the eigenvector for eigenvalue 1, such that ( (A- I)v\ne 0 but such that (A- I)^2v= 0. We can write that as (A- I)((A- I)v)= 0 so that if v is not an eigenvector, (A- I)v must be an eigenvector!

    Since you have found that one eigenvector for eigenvalue 1 is \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}, you want to find \begin{bmatrix}x \\ y \\ z\end{bmatrix} such that
    \begin{bmatrix}3 & 1 & -4 \\ 6 & 1 & -7\\ 6 & 1 & -7\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}

    Such a vector will be the second column you need in T.

    (I once made a total hash of a problem like this and Plato corrected me so I am redeeming myself.)
    Last edited by HallsofIvy; January 17th 2011 at 02:17 PM.
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  9. #9
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    Thank you for this response HallsofIvy, I'd nearly resigned myself to being doomed come tommorow morning, but this is a great explanation. (Except the latex errors =P)
    I now understand this problem much better than I did before.

    Just as a question, would I proceed to make the Matrix (A-I)^2 rREF, to make the equations easier to solve? and Also does nullity(A-I) = geometric multiplicity?

    I totally know what you mean by making a hash of these problems! The numbers are so easy to mix up with your minus signs etc. I'm just going to have to take my sweet time.
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  10. #10
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    I have corrected the LaTex (had to go fix dinner just as I was typing that!). You can solve the equations Av= v, Av= 2v, and (A- I)v= \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} however you like- row reducing the augmented matrix or solving the three equations simultaneously.

    Yes, the "geometric multiplicity" of an eigenvalue, [tex]\lambda[/itex, is the number of independent vectors, v, that satisfy Av= \lamda v which is, of course, the same as (A- \lambda I)v= 0 and so is the "nullity" of A- \lambda I.
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