$$ A = \begin{bmatrix} 0 & 1 & -\sin \theta \\ -1 & 0 & \cos \theta \\ -\sin \theta & \cos \theta & 0 \end{bmatrix} $$ Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $\lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.
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- 2$\begingroup$ "which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$. $\endgroup$amsmath– amsmath2018-08-26 11:54:06 +00:00Commented Aug 26, 2018 at 11:54
- $\begingroup$ Exchanging rows is an elementary row operation. $\endgroup$amd– amd2018-08-26 16:27:41 +00:00Commented Aug 26, 2018 at 16:27
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7 Since you have a $0$ there, exchange the first and the second rows, thereby getting$$\begin{bmatrix}-1&0&\cos\theta\\0&1&\sin\theta\\-\sin\theta&\cos\theta&0\end{bmatrix}.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $\lambda(\cos\theta,\sin\theta,1)$.
answered Aug 26, 2018 at 11:50
José Carlos Santos
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- $\begingroup$ My question is can this be done when I am trying to find the eigenvectors of any matrix? $\endgroup$Harry Touloupas– Harry Touloupas2018-08-26 11:54:40 +00:00Commented Aug 26, 2018 at 11:54
- $\begingroup$ @HarryTouloupas No. But it can be done when the eigenvalue is $0$. $\endgroup$José Carlos Santos– José Carlos Santos2018-08-26 11:55:59 +00:00Commented Aug 26, 2018 at 11:55
- $\begingroup$ @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $\lambda$ you can perform row operations (also exchange of rows) on $A-\lambda I$ in order to find an eigenvector. Here, $\lambda$ is zero, so you do the stuff with $A$. $\endgroup$amsmath– amsmath2018-08-26 11:58:03 +00:00Commented Aug 26, 2018 at 11:58
- $\begingroup$ Also in this example why is $A^k = \vec0$ for every $k>=3$ ? I can see why $A^k = \vec0$ but why do we need the $k>=3$? $\endgroup$Harry Touloupas– Harry Touloupas2018-08-26 12:15:34 +00:00Commented Aug 26, 2018 at 12:15
- $\begingroup$ @HarryTouloupas Because$$A^2=\begin{bmatrix}-\cos ^2(\theta ) & -\sin (\theta ) \cos (\theta ) & \cos (\theta ) \\-\sin (\theta ) \cos (\theta ) & -\sin ^2(\theta ) & \sin (\theta ) \\ -\cos (\theta ) & -\sin (\theta ) & 1\end{bmatrix}\neq0.$$ $\endgroup$José Carlos Santos– José Carlos Santos2018-08-26 12:31:03 +00:00Commented Aug 26, 2018 at 12:31