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Calculate $e^A$ where $$A = \begin{bmatrix}1&0&3\\-1&2&0\\0&1&-1\end{bmatrix}$$

I knew how to do it if it was diagonalizable and real eigenvalues. How can I calculate when the matrix has complex eigenvalues. This matrix $A$ seems to have two conjugate complex evalues.

I know how to calculate $\det \left(e^A\right)$ without knowing $e^A$ as: $$\det \left(e^A\right) = \det \left(e^J\right) = \det \left(e^{\text{tr}J}\right) = \det \left(e^{\text{tr}A}\right) = e^2$$ where $A = CJC^{-1}$ a Jordan form of $A$, but I want to calculate $e^A$.

When I tried to find its eigenvalues to diagonalize it I came to $$\lambda^3-2\lambda^2-\lambda+5 = 0$$ which I couldn't solve.

Also, I would really appreciate some beginner-friendly source references to learn more about problems of this type.

NodeJS
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  • This is missing a little information, can you define what you mean by the matrices $C$ and $J$? –  Jun 10 '21 at 08:37
  • @fundamentalform $J$ is a Jordan block matrix with eigenvalues of $A$ through its diagonal. – NodeJS Jun 10 '21 at 08:43
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    You can compute the matrix exponential using the Cayley-Hamilton theorem, see for example https://math.stackexchange.com/q/724647. – Martin R Jun 10 '21 at 08:55
  • @MartinR But wouldn't my characteristic polynomial be inefficient for this purpose? – NodeJS Jun 10 '21 at 09:24
  • You ask "How can I calculate when the matrix has complex eigenvalues?" Completely analogous to the real-eigenvalues case! $A$ has one real root at approx. $-1.4$ and a pair of complex-conjugate zeros. Diagonalise $A=TDT^{-1}$ & proceed to determine $e^A=Te^DT^{-1}$. – Hanno Jun 10 '21 at 10:13
  • Well, read eigenvalue of $A$, instead of root or zeros. – Hanno Jun 10 '21 at 10:28
  • @Hanno But it was in my exam, so I could not find the eigenvalues. After the exam, I input it into calculator that gave one real and two complex e-values. Also, when the eigenvalues are complex, would not it change the form of $D$. I guess there would $\sin$ and $\cos$ elements in it? I don't have any good source for this stuff unfortunately. – NodeJS Jun 10 '21 at 13:21
  • @Moo Theoretically, I guess yes, but with this particular example I'm not. – NodeJS Jun 10 '21 at 13:22
  • @Moo I would so much appreciate if you could post an answer outlining all the possible cases if you have chance, please. Thank you for your interest. – NodeJS Jun 10 '21 at 15:13
  • This post is related. – Ben Grossmann Jun 14 '21 at 13:57

1 Answers1

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I am not sure what the instructor had in mind for an exam approach, but there are many ways to calculate the matrix exponential, for example Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later.

The closed form eigenvalues / eigenvectors are quite ugly, so we will use numerical values. Because this matrix has three distinct eigenvalues, a typical approach for finding the matrix exponential is using diagonalization.

We have ($P$ are the eigenvector column vectors and $D$ are the corresponding eigenvalues)

$$e^A = P e^ D P^{-1}$$

For the matrix $A$, we have eigenvalues

$$\lambda_{1,2,3} = 1.68721 + 0.889497 i,1.68721 - 0.889497 i,-1.37442$$

The eigenvectors are

$$P = \begin{pmatrix} 1.63173\, -2.11204 i & 2.68721\, +0.889497 i & 1. \\ 1.63173\, +2.11204 i & 2.68721\, -0.889497 i & 1. \\ -1.26346 & -0.374424 & 1. \\ \end{pmatrix}$$

The exponential is given as $e^A = P e^D P^{-1}$

$$\begin{pmatrix} 1.63173\, -2.11204 i & 2.68721\, +0.889497 i & 1. \\ 1.63173\, +2.11204 i & 2.68721\, -0.889497 i & 1. \\ -1.26346 & -0.374424 & 1. \\ \end{pmatrix} \begin{pmatrix} e^{1.68721\, -0.889497 i} & 0. & 0. \\ 0. & e^{1.68721\, +0.889497 i} & 0. \\ 0. & 0. & \frac{1}{e^{1.37442}} \\ \end{pmatrix} \begin{pmatrix} 0.0491893\, -0.169309 i & 0.116796\, +0.160105 i & 0.10588\, -0.153969 i \\ 0.0491893\, +0.169309 i & 0.116796\, -0.160105 i & 0.10588\, +0.153969 i \\ -0.0983786 & -0.233592 & 0.78824 \\ \end{pmatrix} $$

Multiplying these out

$$e^A = \begin{pmatrix} 1.56484 & 3.33455 & 2.90601 \\ -4.30322 & 5.86806 & -3.33455 \\ -1.11152 & 2.08019 & -0.372504 \\ \end{pmatrix}$$

Let's compare this with Wolfram Alpha's result.

Moo
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