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I have to solve the following equation by diagonalization.

$ X' = \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} X$

I was able to determine the complex eigenvalue roots:

$det(A-\lambda I)=0$

$det\begin{bmatrix}1-\lambda & 1\\1 & -1-\lambda\end{bmatrix} =0$

$(1-\lambda)(-1-\lambda)-1 =0$

$\lambda^2 - 2 = 0$

$\lambda_1 = \sqrt2 i$

$\lambda_2 = -\sqrt2 i$

Now I need to find the eigenvectors for diagonalization. However, I get stuck here: there does not seem to be a solution for the system.

$For \lambda_1 = \sqrt2 i : $

$(A-\lambda_1 I)K_1 = 0$

$ \begin{bmatrix}1-\sqrt2 i & 1\\1 & -1-\sqrt2 i\end{bmatrix}$ $\begin{bmatrix} k_1 \\ k_2 \end{bmatrix} =0$

$ \begin{bmatrix}(1-\sqrt2 i)k_1 + k_2\\k_1 + (-1-\sqrt2 i)k_2\end{bmatrix} = 0$

This is where I get stuck. When I attempt to assign values for $k_1$ and $k_2$ to satisfy one of the equations (an example: $k_1 = -(-1-\sqrt2 i)$ and $k_2 = 1$), I end up with $4=0$ for the other equation, which is clearly wrong. This is the first time I ever struggle with finding an eigenvector of $k_1$ and $k_2$ for a system, and I'm wondering if it even has a solution at all.

Is my assignment question wrong? Does this homogeneous linear system of differential equations have a solution? How can I solve for the eigenvectors in this problem?

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    $\begingroup$ The solutions of $\lambda^2 - 2 = 0$ are not $\lambda_1 = \sqrt2 i$ and $\lambda_2 = -\sqrt2 i$. $\endgroup$
    – Did
    Commented Nov 21, 2014 at 22:38

1 Answer 1

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you can write the matrix $$A=\pmatrix{1&1\\1&-1} = \sqrt 2 \pmatrix{\cos \pi/4 & \sin \pi/4\\\sin \pi/4 & -\sin \pi/4 } = \sqrt 2 R$$ where $R$ is a reflection matrix on the line $y = \tan(\pi/8) x$ which has the property $R^2 = I.$

we can find $$\begin{align}e^{At} &= I + tA + \frac{t^2A^2}{2!} + \frac{t^3A^3}{3!}+\cdots \\ &=I + (\sqrt 2 t) R + \frac1{2!}(\sqrt 2 t)^2I+ \frac1{3!}(\sqrt 2 t)^3R + \cdots \\ &=\left(1 + \frac1{2!}(\sqrt 2 t)^2+\cdots \right)I + \left((\sqrt 2 t)+\frac1{3!}(\sqrt 2 t)^3+\cdots\right)R\\ &=\cosh(\sqrt 2 t)I + \sinh(\sqrt 2 t)R\\ &=\cosh(\sqrt 2 t)I + \frac1{\sqrt 2}\sinh(\sqrt 2 t)A\end{align}$$

the solution to the homogeneous equation $$x' = Ax, x(0) = x_0$$ is $$x = e^{At}x_0. $$

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