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$$(\lambda I - A)\vec x=\vec 0$$ This is the $(\lambda I - A)$: $$\begin{bmatrix}\lambda+3 & -1 &1\\7 & \lambda-5 & 1\\-6 & -6 & \lambda+2 \end{bmatrix}$$ The characteristic polynomial is: $$\lambda^3-64$$ My eigenvalues are: $$\lambda_1=4,\\ \lambda_2=-2+\sqrt 3 i,\\\lambda_3=-2-\sqrt 3 i$$ I solved for the real eigenvalue by finding its eigenvector $(0,1,1)^t$,but when I try to do the same with the complex eigenvalues I get that the eigenvector has to be $(0,0,0)^t$ (using a matrix calculator ) which is false according to wolfram because the correct eigenvectors are: $$\begin{pmatrix}\pm\frac{i}{2\sqrt3} \\ \pm\frac{i}{2\sqrt3} \\ 1\end{pmatrix}$$

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  • $\begingroup$ The second eigenvalue $\;1+2\sqrt3\,i\;$, right? $\endgroup$
    – DonAntonio
    Mar 26, 2016 at 13:26
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    $\begingroup$ @PanthersFan92 I can see all the matrix's entries are real and thus the roots of its characteristic polynomial are either real or pairs of conjugated complex non-real numbers: it is impossible you have two real roots and only one complex non-real one for a real polynomial. I haven't checked but I think you're missing $\;i\;$ in the second root. $\endgroup$
    – DonAntonio
    Mar 26, 2016 at 13:34
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    $\begingroup$ The RREF for $-2 + 2i \sqrt{3} $, should be $\left( \begin{array}{ccc} 1 & 0 & \frac{-3 i+7 \sqrt{3}}{6 \left(7 i+\sqrt{3}\right)} \\ 0 & 1 & \frac{-3 i+7 \sqrt{3}}{6 \left(7 i+\sqrt{3}\right)} \\ 0 & 0 & 0 \\ \end{array} \right)$ $\endgroup$
    – Moo
    Mar 26, 2016 at 13:38
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    $\begingroup$ Not that and not that: the eigenvalues are $\;-2\pm2\sqrt3\,i\;$ , just as Moo implied in his comment. $\endgroup$
    – DonAntonio
    Mar 26, 2016 at 13:52
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    $\begingroup$ @PanthersFan92 That's wrong, as noted above: $$z^3=64=64e^{2k\pi i}\implies z_k=4e^{\frac{2k\pi}3}\implies z_0=4\,,\,\color{red}{z_1=-2+2\sqrt3\,i}\,,\,\color{red}{z_2=-2-2\sqrt3\,i}$$ And yes: now you have two complex (conjugated) roots, after you edited the question. When the comments above appeared you only had one. $\endgroup$
    – DonAntonio
    Mar 26, 2016 at 13:57

2 Answers 2

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First note that the correct eigenvalues are: $$ \lambda_1=4 \qquad \lambda_2=-2-2i\sqrt{3}\qquad \lambda_3=-2+2i\sqrt{3} $$ Now , to find the corresponding eigenvector, substitute the eigenvalue in $(\lambda I -A)x=0$ and, solving the homogeneous linear system, find the corresponding eigenspace; any vector in this eigenspace is an eigenvector.

E.g., for $\lambda_3$ we have the system: $$ \begin{cases} (1+2i\sqrt{3})x-y+z=0\\ 7x+(2i\sqrt{3}-7)y+z=0\\ -6x-6y+2i\sqrt{3}z=0 \end{cases} $$ adding the first two equation we find $x=y$ and substituting: $$ -12x+2i\sqrt{3}z=0 \Rightarrow x=\frac{2i\sqrt{3}z}{12}=\frac{iz}{2\sqrt{3}} $$ so the eigenspace is formed by vectors of the form $$ \begin{bmatrix} \frac{iz}{2\sqrt{3}}\\\frac{iz}{2\sqrt{3}}\\z \end{bmatrix} $$ and one possible eigenvector is $$ \begin{bmatrix} \frac{i}{2\sqrt{3}}\\\frac{i}{2\sqrt{3}}\\1 \end{bmatrix} $$

You can do the same for the other eigenvalue.

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Substitute for example$\;\lambda=-2+2\sqrt3\,i\;$ in $\;\det(\lambda I-A)\;$ to obtain the homogeneous system (remember: the system is at most of rank two!):

$$\begin{cases}&(1+2\sqrt3\,i)x-&y&+&z=0\\{}\\&7x-&(7-2\sqrt3\,i)y&+&z=0&\end{cases}\implies(-6+2\sqrt3\,i)x+(6-2\sqrt3\,i)y=0$$$${}$$

$$\implies x=y\implies(1+2\sqrt3\,i)x-x+z=0\implies z=-2\sqrt3\,i$$

and an eigenvector for this eigenvalue is for example

$$\begin{pmatrix}1\\1\\-2\sqrt3\,i\end{pmatrix}$$

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  • $\begingroup$ Are you sure this eigenvector is correct? Emilio gave a different answer $\endgroup$
    – asaf92
    Mar 26, 2016 at 14:32
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    $\begingroup$ @PanthersFan92 Of course: my eigenvector is Emilio's multiplied by a scalar different of zero. Check it. $\endgroup$
    – DonAntonio
    Mar 26, 2016 at 14:35

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