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Math Help - Matrix exponential without series expansion

  1. #1
    Senior Member bkarpuz's Avatar
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    Matrix exponential without series expansion

    Dear MHF members,

    my problem is explained below.
    __________________________________________________
    Problem. Compute the matrix exponential of the matrix
    A:=<br />
\left(<br />
  \begin{array}{rrrr}<br />
    1 & 0 & -1 & 1 \\<br />
    0 & 1 & 1 & 0 \\<br />
    0 & 0 & 1 & 0 \\<br />
    0 & 0 & 1 & 0 \\<br />
  \end{array}<br />
\right).
    ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
    I can compute the matrix exponential by using the series expansion
    \mathrm{e}^{At}=\mathrm{I}+\displaystyle\sum_{\ell  =1}^{\infty}\displaystyle\frac{1}{\ell!}A^{\ell}t^  {\ell},
    where
    A^{n}=<br />
\left(<br />
  \begin{array}{rrrr}<br />
    1 & 0 & -1 & 1 \\<br />
    0 & 1 & n & 0 \\<br />
    0 & 0 & 1 & 0 \\<br />
    0 & 0 & 1 & 0 \\<br />
  \end{array}<br />
\right) for n\in\mathbb{N}.

    However, I could not figure out how to compute it by using the eigenvalues.
    I would be glad if you show me the solution in this way.

    Thanks.
    bkarpuz
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    You can find the canonical form of Jordan J for A and a non singular matrix P such that P^{-1}AP=J . Then, e^{tA}=Pe^{tJ}P^{-1} .
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  3. #3
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by FernandoRevilla View Post
    You can find the canonical form of Jordan J for A and a non singular matrix P such that P^{-1}AP=J . Then, e^{tA}=Pe^{tJ}P^{-1}.
    Actually, this is the part I am having problems since it has repeated eigenvalues.
    Do you mind explaining me?

    Thanks.
    bkarpuz
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  4. #4
    MHF Contributor

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    Yes, and so the matrix may not be diagonalizable. But the crucial part is not the number of eigenvalues but the number of independent eigenvectors. Here, 0 is a single eigenvalue and 1 is triple eigenvalue. An eigenvector corresponding to eigenvalue 0 (any multiple of <1, 0, 0, -1> is such an eigenvalue) is independent of any other eigenvalues but the eigenvalue 1 may have 3, 2, or only 1 independent eigenvector. In fact, looking for eigenvectors corresponding to eigenvalue 1 means looking for solutions to
    \begin{pmatrix}1 & 0 & -1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}=\begin{pmatrix}x- z+ w\\ y+ z \\ z \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}
    So we must have x- z+ w= x, y+ z= y, z= z, and z= w y+ z= y gives immediately z= 0 so then w= 0. x- z+ w= x becomes x= x and y+ z= 7 becomes y= y both of which are satisfied by all x and y. That is, any eigenvector corresponding to eigenvalue 1 is of the form <x, y, 0, 0>= <x, 0, 0, 0>+ <0, y, 0, 0>= x<1, 0, 0, 0>+ y<0, 1, 0, 0>. That is, <1, 0, 0, 0> and <0, 1, 0, 0> are independent eigenvectors corresponding to eigenvalue 1. Because there are only 3 independent eigenvectors and not 4, we cannot "diagoalize" this matrix, but we can write it in "Jordan Normal form' as Fernando Revilla suggested. To do that, we need to find a "generalized eigenvector"- a vector, v, such that (A- \lambda I)v\ne 0 but (A- \lambda I)^2v= 0. That, of course, is the same as saying (A- \lambda)((A- \lambda)v))= 0 which, in turn, means that (A- \lambda I)v is an eigenvector. That is, we must have
    \begin{pmatrix}1-1 & 0 & -1 & 1 \\ 0 & 1-1 & 1 & 0 \\ 0 & 0 &  1-1 & 0 \\ 0 & 0 & 1 & 1-1\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}
    must be equal to either \begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} or \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}.

    The first of those is impossible but the second is satisfied by \begin{pmatrix}0 \\ 1 \\ 1 \\ 1\end{pmatrix}.

    Constructing a matrix, P, having those vectors as columns, say,
    P= \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix}
    then P^{-1}AP will be the Jordan Normal form.

    (That's the idea. The calculations were done quickly and I don't guarentee them.)
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