Question

Consider a$\mathbf{2}\mathbf{\times }\mathbf{2}$matrix$\mathit{A}$with eigenvalues$\mathbf{\pm }\mathit{\pi }\mathit{i}$. Let $\overrightarrow{\mathbf{v}}\mathbf{+}\mathit{i}\overrightarrow{\mathbf{w}}$be an eigenvector of A with eigenvalue . Solve the initial value problem $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$with${\overrightarrow{\mathbf{x}}}_{\mathbf{0}}\mathbf{=}\overrightarrow{\mathbf{w}}$.

Draw the solution in the accompanying figure. Mark the vectors$\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{,}\overrightarrow{\mathbf{x}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}\mathbf{,}\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{,}$and$\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$.

Short answer

The solution is$\overrightarrow{x}\left(1\right)=-\overrightarrow{w},\overrightarrow{x}\left(2\right)=\overrightarrow{w},\overrightarrow{x}\left(\frac{1}{2}\right)=\overrightarrow{v},\overrightarrow{x}\left(0\right)=\overrightarrow{w}$. The vectors are $\overrightarrow{x}\left(0\right),\overrightarrow{x}\left(\frac{1}{2}\right),\overrightarrow{x}\left(1\right)$and $\overrightarrow{x}\left(2\right)$.

Step-by-step solution

Definition of the theorem

Continuous dynamical system with eigenvalues$\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$

Consider the linear system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$where$\mathit{A}$is a real$\mathbf{2}\mathbf{\times }\mathbf{2}$matrix with complex eigenvalues$\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$(and$\mathit{q}\mathbf{\ne }\mathbf{0}$).

Consider an eigenvector$\overrightarrow{\mathbf{v}}\mathbf{+}\mathit{i}\overrightarrow{\mathbf{w}}$with eigenvalue$\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$.

Then$\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{p}\mathbf{t}}\mathit{S}\mathbf{\left[}\begin{array}{cc}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}& \mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}\\ \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}& \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}\end{array}\mathbf{\right]}{\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$where$\mathit{S}\mathbf{=}\mathbf{\left[}\overrightarrow{\mathbf{w}}\overrightarrow{\mathbf{v}}\mathbf{\right]}$. Recall that${\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$is the coordinate vector of${\overrightarrow{\mathbf{x}}}_{\mathbf{0}}$with respect to basis$\overrightarrow{\mathbf{w}}\mathbf{,}\overrightarrow{\mathbf{v}}$.

Explanation of the solution

Consider a$2\times 2$matrix$A$with eigenvalues$\pm \pi i$. Let$\overrightarrow{v}+i\overrightarrow{w}$be an eigenvector of$A$with eigenvalue$\pi i$.

Now, consider the system as follows.

$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$with the initial condition${\overrightarrow{x}}_0=\overrightarrow{w}$.

If$A$has eigenvalue$p\pm iq(q\ne 0)$and$\overrightarrow{v}+i\overrightarrow{w}$is the eigenvector of$p\pm iq$, then the solution as follows.

$\begin{array}{l}\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0\\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}\end{array}$

Substitute the value$0$for$t$in$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}$as follows.

$\begin{array}{c}\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi \left(0\right)\right)& -sin\left(\pi \left(0\right)\right)\\ sin\left(\pi \left(0\right)\right)& cos\left(\pi \left(0\right)\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(0\right)& -sin\left(0\right)\\ sin\left(0\right)& cos\left(0\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \overrightarrow{x}\left(0\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \overrightarrow{x}\left(0\right)=\overrightarrow{w}\end{array}$

$\begin{array}{l}\overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \overrightarrow{x}\left(0\right)=\overrightarrow{w}\end{array}$

Substitute the value$\frac{1}{2}$for$t$in$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}$as follows.

$\begin{array}{c}\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi \left(\frac{1}{2}\right)\right)& -sin\left(\pi \left(\frac{1}{2}\right)\right)\\ sin\left(\pi \left(\frac{1}{2}\right)\right)& cos\left(\pi \left(\frac{1}{2}\right)\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\frac{\pi }{2}\right)& -sin\left(\frac{\pi }{2}\right)\\ sin\left(\frac{\pi }{2}\right)& cos\left(\frac{\pi }{2}\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\overrightarrow{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]}^{-1}\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{c}\overrightarrow{x}\left(\frac{1}{2}\right)=\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ 0\end{array}\right]\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(\frac{1}{2}\right)=\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\end{array}$

Simplify further as follows.

$\overrightarrow{x}\left(\frac{1}{2}\right)=\overrightarrow{v}$

Simplification of the solution

Substitute the value$1$for$t$in$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}$as follows.

$\begin{array}{c}\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(1\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi \left(1\right)\right)& -sin\left(\pi \left(1\right)\right)\\ sin\left(\pi \left(1\right)\right)& cos\left(\pi \left(1\right)\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(1\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi \right)& -sin\left(\pi \right)\\ sin\left(\pi \right)& cos\left(\pi \right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\overrightarrow{x}\left(1\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \overrightarrow{x}\left(1\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\overrightarrow{x}\left(1\right)=\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(1\right)=\left[\begin{array}{c}-w_1\\ -w_2\end{array}\right]\\ \overrightarrow{x}\left(1\right)=-\overrightarrow{w}\end{array}$

Substitute the value$2$for$t$in$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\pi t\right)& -sin\left(\pi t\right)\\ sin\left(\pi t\right)& cos\left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}$as follows.

$\begin{array}{c}\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(\pi t\right)& -\sin \left(\pi t\right)\\ \sin \left(\pi t\right)& \cos \left(\pi t\right)\end{array}\right]S^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(2\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(\pi \left(2\right)\right)& -\sin \left(\pi \left(2\right)\right)\\ \sin \left(\pi \left(2\right)\right)& \cos \left(\pi \left(2\right)\right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ \overrightarrow{x}\left(2\right)=\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(2\pi \right)& -\sin \left(2\pi \right)\\ \sin \left(2\pi \right)& \cos \left(2\pi \right)\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\overrightarrow{x}\left(2\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\overrightarrow{w}& \overrightarrow{v}\end{array}\right]}^{-1}\overrightarrow{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \overrightarrow{x}\left(2\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\overrightarrow{x}\left(2\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(2\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \overrightarrow{x}\left(2\right)=\overrightarrow{w}\end{array}$

Diagram representation of the solution

The vectors $\overrightarrow{x}\left(0\right),\overrightarrow{x}\left(\frac{1}{2}\right),\overrightarrow{x}\left(1\right)$and $\overrightarrow{x}\left(2\right)$is pointed in the figure 1 as follows.

Thus, the vectors $\overrightarrow{x}\left(0\right),\overrightarrow{x}\left(\frac{1}{2}\right),\overrightarrow{x}\left(1\right)$and$\overrightarrow{x}\left(2\right)$are shown in figure 1.