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Math Help - Matrix - Jordan Normal/Canonical Form

  1. #1
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    Matrix - Jordan Normal/Canonical Form

    QUESTION:

    Find A^k when:

    A =  \begin{bmatrix} -1 & -2 \\ 2 & 3 \end{bmatrix}

    __________________________________________________ ___

    The eigenvalue (of multiplicity 2) is  x = 1

    I've tried putting A into Jordan-Normal/Canonical form, but I find that the matrix (A- x I)^2 = 0 , and I don't know what to do.

    Please help.
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  2. #2
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    Hello, WWTL@WHL!

    Find A^k when: .  <br />
A \:= \: \begin{bmatrix} \text{-}1 & \text{-}2 \\ 2 & 3 \end{bmatrix}

    I see no choice but to crank out a few powers and seek a pattern . . .

    . . A \;=\;\begin{bmatrix}\text{-}1&\text{-}2\\2&3\end{bmatrix}

    . . A^2 \;=\;\begin{bmatrix}\text{-}3 & \text{-}4 \\ 4 & 5 \end{bmatrix}

    . . A^3 \;=\;\begin{bmatrix}\text{-}5&\text{-}6\\6&7\end{bmatrix}

    . . A^4 \;=\;\begin{bmatrix}\text{-}7&\text{-}8\\8&9\end{bmatrix}

    . . A^5 \;=\;\begin{bmatrix}\text{-}9&\text{-}10\\10&11\end{bmatrix}



    I think we've found the pattern . . .

    We have: . A^k \:=\:\begin{bmatrix}a&b\\c&d\end{bmatrix}

    . . where: a is the negative of the k^{th} odd number: . -2k+1

    . . . . . . . b is one less than a\!:\;\;-2k

    . . . . . . . c is the positive value of b\!:\;\;2k

    . . . . . . . d is one more than c\!:\;\;2k+1


    Therefore: . A^k \;=\;\begin{bmatrix}\text{-}2k+1 & \text{-}2k \\ 2k & 2k+1 \end{bmatrix}

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  3. #3
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    here is a more general way.

     det \begin{bmatrix} -1-x & \text{-}2 \\ 2 & 3-x \end{bmatrix}<br />
=(-1-x)(3-x)+4=x^2-2x+1=(x-1)^2
    so we need to find the canonical form. we can see that the matrix has only one eigenvector for 1, so the jordan form is
    J=P^{-1}AP=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}
    to find P we can write P=[u,v] where u,v are vectors of size two and
    AP=PJ \rightarrow Av=v, Au=v+u
    so v is an eigenvector, for example (1,-1), and to find u solve the equation (A-I)u=v.
    now you have A^k = (PJP^{-1})^k=PJ^k P^{-1}
    notice that this is a general way to find powers of matrices that have the jordan form. because this is used a lot, it is good to know what happens when you raise a jordan matrix to the power of k. to make it easier, write J=dI+Z where the d is the eigenvector (in this case 1) and the Z matrix is a zero matrix with ones only above the main diagonal. in this case Z would just be \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} and then you have
    J^k = (dI+Z)^k = \sum \binom{k}{i} d^{k-i}Z^i
    and the thing about raising Z to the power of i is just just moving the "ones" i times to the top right corner, and here, because Z is only 2x2 matrix, moving the "ones" one time will take them out of the matrix, meaning Z^2=0 and because d=1 you get J^k=\binom{k}{0}I+\binom{k}{1}Z=\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}
    this method works on the more general cases. for example, if you have more than one jordan block, then you can do this for every block separately.
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  4. #4
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    Quote Originally Posted by WWTL@WHL View Post
    QUESTION:

    Find A^k when:

    A =  \begin{bmatrix} -1 & -2 \\ 2 & 3 \end{bmatrix}

    __________________________________________________ ___

    The eigenvalue (of multiplicity 2) is  x = 1

    I've tried putting A into Jordan-Normal/Canonical form, but I find that the matrix (A- x I)^2 = 0 , and I don't know what to do.

    Please help.
    Yes, of course (A- I)^2= 0 because every matrix satisfies its own characteristic equation.

    First look for a eigenvectors for \lambda= 1:
    \begin{bmatrix}-1 & -2 \\ 2 & 3\end{matrix}\begin{matrix} x \\ y\end{matrix}= \begin{matrix} x \\ y\end{bmatrix}
    which gives the equations -x- 2y= x and 2x+ 3y= y. Those both reduce to y= -x so <1, -1> is an eigenvector but we don't find a second. But because (A- I)^2 v= 0 for all vectors v and there are some such that (A- I)v is not 0, there must be v such that (A- I)((A-I)v)= 0 which means (A-I)v is an eigenvalue, a multiple of <1, -1>. That is, look for v such that (A- I)v= \begin{bmatrix}-2 & -2 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}1 \\ -1\end{bmatrix}
    That gives the equations -2x-2y= 1 and 2x+ 2y= -1 which are, again, equivalent. But taking x= 1 we get y= -3/2 so <1, -3/2> is a "generalized eigenvalue". Create the matrix \begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix} with those vectors as columns and find its inverse, \begin{bmatrix}3 & 2 \\ -2 & -2\end{matrix}

    Now we have \begin{bmatrix}3 & 2 \\ -2 & -2\end{bmatrix}\begin{bmatrix}-1 & -2 \\ 2 & 3\end{bmatrix}\begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}, the "Jordan Normal Form" for A.

    That is, P^{-1}AP= N so A= PNP^{-1} and A^n= PN^nP^{-1}

    And, of course, N^n= \begin{bmatrix}1 & n \\ 0 1 so to find A^n, you just need to calculate \begin{bmatrix}1 & 1 \\ -1 & -3/2\end{bmatrix}\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ -2 & -2\end{bmatrix}.
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