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Usng the spectral decompostion of diagonalizable matrix

  1. Oct 1, 2012 #1

    syj

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    1. The problem statement, all variables and given/known data
    Find the spectral decompostion of
    $$
    A=

    \begin{matrix}
    1 & 0 & 0\\
    -1 & 1 & 1 \\
    -1 & 0 & 2
    \end{matrix}
    $$
    and use this to find $$ 2^{A} $$

    2. Relevant equations



    3. The attempt at a solution
    I have found the eigenvalues to be : $$ \lambda_{1}=1 \text{ and } \lambda_{2}=2 $$

    My textbook (matrices and linear transformations, Cullen) uses lagrange polynomials to find the spectral projectors:
    $$
    h_{1}(x)=\frac{x-2}{1-2}=-(x-2)\\
    h_{2}(x)=\frac{x-1}{2-1}=(x-2)\\

    E_{1}=h_{1}(A)\\

    E_{2}=h_{2}(A)\\

    \therefore A=E_{1}+2E_{2}

    $$
    However I now need to use this to solve for $$ 2^{A} $$
    I think I need to write $$ 2^{x} $$ as an infinite series, but I am not sure how to do this.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    You know how to write [itex]e^A[/itex] as an infinite series, right? [itex]2^A=e^{A \log 2}[/itex]
     
  4. Oct 1, 2012 #3

    syj

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    The textbook I am using says:
    [tex]
    f(A)=\sum_{i=0}^{t}f(\lambda_{i})E_{i}
    [/tex]

    I still don't understand how to put everything together to get the answer. :(
     
  5. Oct 1, 2012 #4

    syj

    User Avatar

    After much puzzling, I believe I have the answer:

    [tex]
    2^{A}=2E_{1}+2^{2}E_{2}
    [/tex]

    where I have found
    [tex]
    E_{1}=\begin{bmatrix} 1&0&0\\1&1& \-1\\1&0&0\end{bmatrix}
    [/tex]

    [tex]
    E_{2}=\begin{bmatrix}0&0&0\\-1&0&1\\-1&0&1\end{bmatrix}
    [/tex]

    so I have
    [tex]
    2^{A}=\begin{bmatrix} 2&0&0\\\-2&2&2\\\-2&0&4\end{bmatrix}
    [/tex]
     
    Last edited: Oct 1, 2012
  6. Oct 1, 2012 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    Why did it require much puzzling? You gave the answer in a previous post:
    [tex] f(A) = f(\lambda_1) E_1 + f(\lambda_2) E2,[/tex]
    and f(x) = 2x.

    RGV
     
  7. Oct 1, 2012 #6

    syj

    User Avatar

    ha ha , it required puzzling, because I knew that
    [tex]
    p(x)=\sum_{i=1}^{n}p(\lambda_{i})h_{i}(x)
    [/tex]
    would have to be such that
    [tex]
    p(A)=2^{A}
    [/tex]
    but because of a silly mistake of writing a 0 instead of a 1, I was not getting that conclusion and so thought all my work was incorrect.
     
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