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$$x'_{1}=x_{1}-x_{3}$$

$$x'_{2}=4x_{1}-3x_{2}-x_{3}$$

$$x'_{3}=x_{1}+x_{3}$$

What i tried

I first convert it to a system of 3x3 matrix

$$ A=\begin{bmatrix} 1 & 0 & -1 \\ 4 & -3 & -1 \\ 1 & 0 & 1 \end{bmatrix} $$

Then i compute out the eigenvalue by forming the characteristic polynomial

$(1-{\lambda})({-3-\lambda})(1-{\lambda})=0$

Solving gives

${\lambda}=1$,${\lambda}=-3$ and ${\lambda}=2$ for the eigenvalue. It is from this part onward that im stuck.

I know i need to find eigenvectors, so i substitute each eigenvalue into the

characteristic polynomial say for ${\lambda}=0$, i got

$$ A=\begin{bmatrix} 1 & 0 & -1 \\ 4 & -3 & -1 \\ 1 & 0 & 1 \end{bmatrix} $$

I know i have to solve the following system of equations to get the eigenvector, but im unsure of how to proceed. Could anyone please explain. Thanks

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  • 2
    $\begingroup$ You went from: $(1- \lambda)(1- \lambda)(-3- \lambda)=0$ to $\lambda =0$? $\endgroup$ – abcdef Apr 1 '15 at 7:28
  • 1
    $\begingroup$ Sorry should be $1$ $\endgroup$ – ys wong Apr 1 '15 at 7:39
  • $\begingroup$ ...so then you shouldn't solve $AX=0$, but $(A-I)X=0$. $\endgroup$ – Hans Lundmark Apr 1 '15 at 7:53
  • $\begingroup$ Note also that the $x_1,x_3$ subsystem is isolated from $x_2$, so you could solve this 2-dimensional system first. This then shows you that you got the characteristic polynomial and the eigenvalues wrong. There is a complex pair of conjugate eigenvalues. $\endgroup$ – LutzL Apr 1 '15 at 9:39
  • $\begingroup$ Do i have to find the inverse of each matrix $A$ for each eigenvalue in order to solve the systems of equations? $\endgroup$ – ys wong Apr 1 '15 at 11:20

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