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The question is

Find $\exp(D)$ where $D = \begin{bmatrix}5& -6 \\ 3 & -4\end{bmatrix}. $

I am wondering does finding the $\exp(D)$ requires looking for the canonical form... Could someone please help?

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Hint:

  • Write the Jordan Normal Form (it is diagonalizable) with unique eigenvalues.
  • $e^{D} = P \cdot e^J \cdot P^{-1}$

The Jordan Normal Form is:

$$A = P J P^{-1} = \begin{bmatrix}1&2 \\ 1 & 1\end{bmatrix}~\begin{bmatrix}-1&0 \\0 & 2\end{bmatrix}~\begin{bmatrix}-1&2 \\ 1 & -1\end{bmatrix}$$

Now use the above formula to find the exponential of $D$.

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  • $\begingroup$ is the answer going to be the following $\exp(D)$ where $D = \begin{bmatrix}2e^2 - e^{-1} & -2e^2+2e^{-1} \\ e^2-e^{-1} & -e^2+2e^{-1}\end{bmatrix}. $ ? $\endgroup$ – afsdf dfsaf Jan 30 '14 at 3:25

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