Struggling to simplify the matrix exponential of the following matrix: $$A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ It's trivial to observe the eigenvalues are $a±bi$, but the algebra gets pretty involved if I can't figure out how to apply that knowledge. Is there a trick to simplify this exercise?
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1You have found the eigenvalues, cool. Can you find corresponding eigenvectors? – Arthur Apr 04 '20 at 15:17
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a+bi has [i; 1] and the other has [-i; 1] – cryptograthor Apr 04 '20 at 15:18
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1Cool. So can you diagonalize the matrix, then? – Arthur Apr 04 '20 at 15:19
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by T^-1 * A * T where T is [v_1 v_2], v_i are the eigenvectors? Yes – cryptograthor Apr 04 '20 at 15:22
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1Yes. But since you're working with $A$ in the first place, it might be a better idea to write it the other way, as $A = TDT^{-1}$ instead, where $D$ is diagonal with eigenvalues as entries. – Arthur Apr 04 '20 at 15:24
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Are you suggesting that I could take $e^A = e^{TDT^-1} = Te^DT^{-1}$ ? – cryptograthor Apr 04 '20 at 15:27
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1That's exactly what I am suggesting. It really is that easy (although you need a small argument by expanding the power series definition of $e^{TDT^{-1}}$ to show that it is indeed equal to $Te^DT^{-1}$). And taking exponentials of diagonal matrices are easy. – Arthur Apr 04 '20 at 15:28
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Yes. Diagonalisation is a good way to compute the matrix exponential. – Parcly Taxel Apr 04 '20 at 15:29
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1I claim to know how to prove that. Thanks @Arthur! – cryptograthor Apr 04 '20 at 15:41
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@ParclyTaxel So it seems! But, perhaps just as interestingly, how many poor ways can you think of? Algebra for instance, seems ill suited, but possible, after juggling out matrix multiplications and power series – cryptograthor Apr 04 '20 at 15:43
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No need to diagonalize. See the complex eigenvalues case of this answer. – amd Apr 04 '20 at 19:38
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Related – Rodrigo de Azevedo Mar 29 '25 at 13:52
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Taking advantage of the Cayley-Hamilton theorem as described in this answer, we set $$G=A-aI=\begin{bmatrix}0&-b\\b&0\end{bmatrix}$$ to obtain $$e^{at}\exp(tG) = e^{at}\left((\cos{bt}) I+\frac{\sin{bt}}b G\right) = \begin{bmatrix}e^{at}\cos{bt}&-e^{at}\sin{bt}\\e^{at}\sin{bt}&e^{at}\cos{bt}\end{bmatrix}.$$
If you note the isomorphism between matrices of the form in your question and complex numbers, you might even guess that this would be the solution from the corresponding identity $e^{t(a+ib)} = e^{at}(\cos{bt}+i\sin{bt})$.
amd
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Only thing I would add is that $\frac1{ib} G$ is involutory. – Rodrigo de Azevedo Mar 29 '25 at 13:48
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Hint: $e^A=Pe^DP^{-1}$
where $D=diag(a+ib,a-ib)$ is diagonal matrix and $P$ is the matrix having the eigenvectors $v_1$ and $v_2$ as its columns. Needless to mention here that $e^D=diag(e^{a+ib}, e^{a-ib})$
Nitin Uniyal
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