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I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.

Here is how I find eigenvalues: $$\begin{align*} \det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} \Bigg)\\ &= \det \Bigg(\begin{bmatrix} -\lambda&1&0 \\ 0&-\lambda&1 \\ -1&0&-\lambda \end{bmatrix} \Bigg)\\ &= -\lambda^3 - 1\\ \therefore \lambda =& -1 \end{align*}$$

Using eigenvalue that I found ($-1$), I want to find eigenvectors: $$\begin{align*} (A - \lambda I)\vec{V} =& 0\\ \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\Bigg) \begin{bmatrix}x\\y\\z \end{bmatrix} =& \begin{bmatrix}0\\0\\0\end{bmatrix}\\ \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\ \begin{bmatrix} x+y \\ y+z \\ -x+z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\ \end{align*}$$

But what I should do from now? What is really the eigenvectors? Does this means that I have unlimited eigenvectors and any number that satisfies three equations can be eigenvectors?

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  • $\begingroup$ There are at most three linear independent eigenvectors (but there could be fewer if one of the eigenvalues is $0$). $\endgroup$ – Jared May 14 '14 at 23:25
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    $\begingroup$ Use $\lambda = -1$ (instead of your current $-\lambda$ along the diagonal of $\lambda I$, substitute $-1$ for each diagonal entry) in the last part to get a system of three variables $x,y,z$ and solve this for the eigenvector corresponding to the eigenvalue $-1$ that you found. $\endgroup$ – afedder May 14 '14 at 23:30
  • $\begingroup$ Good, now solve this to obtain the eigenvector. Solving systems of simultaneous equations is something taught in an elementary algebra course, but you can use linear algebra here and just row-reduce the matrix $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}$ $\endgroup$ – afedder May 14 '14 at 23:44
  • $\begingroup$ Note that there may be more than $1$ solution, but you can find the general eigenvector - in fact, there are since subtracting the second equation from the first equation and adding the result to the third equation yields $0=0$. $\endgroup$ – afedder May 14 '14 at 23:45
  • $\begingroup$ All of the distinct eigenvectors should be orthogonal (i.e. linearly independent). Where this gets tricky is when you have a multiplicity--especially if you have an eigenvalue of $0$ (but you don't have that in this case). It's pretty trivial to show that once you have an eigenvector $\vec{v}_1$ that $\alpha \vec{v}_1$ ($\alpha \neq 0$) is also an eigenvector (but those two will not be orthogonal thus they are not distinct eigenvectors). $\endgroup$ – Jared May 14 '14 at 23:49
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Since your characteristic equation is:

$$ \lambda^3 = -1 \rightarrow \lambda = e^{\pi i + \frac{2n\pi}{3}i} $$ and gives three distinct eigenvalues, there are exactly three eigenvectors only one of which has eigenvalue $\lambda = -1$.

$$ \begin{bmatrix} 1 & 1 & 0 \\ 0 &1& 1 \\ -1 &0 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 0 &1& 1 \\ 0 &1 & 1 \end{bmatrix} $$

Now the last two are degenerate (as we would expect) which gives:

$$ y = -z \\ x = -y = z \\ (z, -z, z) \rightarrow (1, -1, 1) $$

So $\left\langle1, -1, 1\right\rangle$ or $\left\langle \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$ is the only eigenvector for $\lambda = -1$.

By only eigenvector, I mean that all eigenvectors for $\lambda = -1$ will be scalar multiples of the above.

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Well, the polynomial $\lambda^3+1$ has two more (complex) roots, which means a rotation in a $2$ dimensional subspace.

In your last equation substitute $\lambda=-1$ and, say, $x=1$ to find one eigenvector.
(You are right: there are infinitely many eigenvectors if there is one as they always form a subspace.)

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For any scalar $k$, if $v$ is an eigenvector for the eigenvalue $\lambda$ ($Av=\lambda v$) then so is $kv$ ($A(kv)=kAv=k\lambda v=\lambda(kv)$). The last line of your set of matrix equations is the homogeneous system $x+y=0$, $y+z=0$, $-x+z=0$. This system is not linearly independent, but that's OK. Doing a bit of solving gives you $x=z$ and $y=-z$, so $(x,y,z)=(z,-z,z)=z(1,-1,1)$. Yes, you get infinitely many eigenvectors, but they are all scalar multiples of eigenvector $(1,-1,1)$. This is normal -- you actually get a whole subspace of eigenvectors for any eigenvalue, with the dimension of the subspace corresponding to the multiplicity of the root in the characteristic polynomial. The vector $(1,-1,1)$ provides a basis for the eigenspace associated with your eigenvalue $\lambda=1$. (You should also note that, in your example, you also get two complex roots -- if your base field is $\mathbb C$ then you get eigenspaces for those values, too.) [Edit: fixed a typo]

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We could put the solution in a cleaner form:

Let $\lambda =-1$ be our eigenvalue. Then we have $$ \begin{align} M&=A-\lambda I\\&= \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \end{align} $$ Let the corresponding eigenvector be $\overrightarrow{v}$: $$ \overrightarrow{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix} $$ such that $M\overrightarrow{v}=\overrightarrow{0},\ \overrightarrow{v}\ne0$. Then we have $$ v_1+v_2=0\\ v_2+v_3=0\\-v_1+v_3=0 $$ Solving the system of linear equations, we have: $$ ker(M)=\left\{\;t\begin{pmatrix}1\\-1\\1\end{pmatrix},\; t\in\mathbb{R}\backslash \left\{0\right\} \right\} $$ Hence the corresponding eigenvector is $$ \overrightarrow{v}=\begin{bmatrix}1\\-1\\1\end{bmatrix} $$

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