How to find matrix exponential $e^A$

Question

I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$

and I have to find $e^A$

I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$

so substracting $\lambda_1 = i$ from the matrix's diagonal I got:

$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix}$$

and therefore. to find eigenvector I have to solve the system:

$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

so the first eigenvector is $h_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$ and the second one is $h_2 = \begin{pmatrix}1 \\ -i\end{pmatrix}$

so the general solution is $$x(t) = C_1e^{it}\begin{pmatrix} 1 \\i\end{pmatrix} + C_2e^{-it}\begin{pmatrix} 1 \\-i\end{pmatrix}$$

I know that now I have to solve two Cauchy's problems for the standard basis $\mathbb{R}^2$ with vectors $v_1 = (1, 0)$ and $v_2 = (0,1)$ But I do not know how to approach it for complex numbers

Answer 1

Let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix};$$its columns are the eigenvectors that you found. Then$$T^{-1}.A.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}.$$Therefore,$$T^{-1}.A^n.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}^n=\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}$$and so$$T^{-1}.e^A.T=\sum_{n=0}^\infty\frac1{n!}\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}=\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.$$Therefore,\begin{align}e^A&=T.\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.T^{-1}\\&=\begin{pmatrix}\frac{e^i+e^{-i}}2&\frac{e^i-e^{-i}}{2i}\\-\frac{e^i-e^{-i}}{2i}&\frac{e^i+e^{-i}}2\end{pmatrix}\\&=\begin{pmatrix}\cos 1&\sin 1\\-\sin1&\cos1\end{pmatrix}.\end{align}

Answer 2

Like here For any $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^{n}.$ Employing the Identification $$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right)=A.$$

we get, $$\color{red}{A^n = i^n ~~~and~~~~ e^A =\sum_{n=0}^{\infty} \frac{i^n}{n!} = e^i = \cos 1+i \sin 1 = \begin{pmatrix}\cos 1&\sin 1\\-\sin1&\cos1\end{pmatrix}.}$$

Answer 3

If the matrix have real coefficients then it complex eigenvalues comes paired, that is, if $\lambda$ is an eigenvalue of $A$ then it is also $\bar\lambda$. For a $2\times 2$ matrix this mean that exists some change of basis $S$ such that $SAS^{-1}=\left[\begin{smallmatrix}\alpha&-\beta\\\beta&\alpha\end{smallmatrix}\right]$, where $\lambda=\alpha+i\beta$.

Then it can be shown that the matrices $B=\left[\begin{smallmatrix}\alpha&0\\0&\alpha\end{smallmatrix}\right]$ and $C=\left[\begin{smallmatrix}0&-\beta\\\beta&0\end{smallmatrix}\right]$ commute, then, from the definition of the exponential map, it can be checked that

$$e^A=S^{-1}e^{SAS^{-1}}S=S^{-1}e^{B+C}S=S^{-1}e^{\alpha I}\cdot e^CS=e^\alpha S^{-1}\begin{bmatrix}\cos\beta&-\sin\beta\\\sin\beta&\cos\beta\end{bmatrix}S$$

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This result can be extended to find what is called the extended normal Jordan form of any real-valued matrix, what make easy the computation of it exponential. Take a look here.

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ADDITION: in the simple case that $A=\left[\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right]$ we can use directly the definition of the exponential map noticing that $A^2=-I$, $A^3=-A$, $A^4=I$ and $A^5=A$. Thus

$$\begin{align}e^A:&=\sum_{k=0}^\infty\frac{A^k}{k!}\\&=\sum_{k=0}^\infty (-1)^k\frac{I}{(2k)!}+\sum_{k=0}^\infty(-1)^k\frac{A}{(2k+1)!}\\&=\begin{bmatrix}\sum_{k=0}^\infty(-1)^k\frac1{(2k)!}&0\\0&\sum_{k=0}^\infty(-1)^k\frac1{(2k)!}\end{bmatrix}+\begin{bmatrix}0&\sum_{k=0}^\infty(-1)^k\frac1{(2k+1)!}\\\sum_{k=0}^\infty(-1)^k\frac{-1}{(2k+1)!}&0\end{bmatrix}\\&=\begin{bmatrix}\cos 1&0\\0&\cos 1\end{bmatrix}+\begin{bmatrix}0&\sin 1\\\sin(-1)&0\end{bmatrix}\\&=\begin{bmatrix}\cos 1&\sin 1\\-\sin 1&\cos 1\end{bmatrix}\end{align}$$