How do i find eigenvectors for a $3\times 3$-matrix when eigenvalues are mixed complex or real?

My task is this;

Find the eigenvalues and eigenvectors for the matrix: $A = \begin{pmatrix}2 & 1 & -3\\4 & 2 & 3\\0 & 0 & 1\end{pmatrix}$.

My work so far is this:

Apply the lemma that states $\lambda$ is and eigenvalue for $n\times n$-matrix M iff $det(\lambda I_n - M) = 0$.

We get (developing $A$ along row 3 in the determinant step):

$\lambda I_3 - A = \begin{pmatrix}\lambda - 2& -1 & 3\\-4&\lambda - 2&-3\\0& 0&\lambda - 1\end{pmatrix} \to det(A) = (\lambda - 1)\begin{vmatrix}\lambda - 2& -1\\-4 & \lambda - 2\end{vmatrix} =\\[-10mm]$ $(\lambda - 1)((\lambda - 2)^2 - 4)= \lambda(\lambda - 1)(\lambda - 4)$.

Which gives us $\lambda = \{0, 1, 4\}$ which in this case were real, but i'm supposed to solve for mixed complex solutions aswell (recall that complex solutions occur in pairs and that we will get three solutions for a third degree polynomial from the determinant of a $3\times 3$).

We should get three equations: $$$$Av_1 = 0v_1\\ Av_2 = v_2\\ Av_3 = 4v_3$$$$

Now i'm not sure how to determine the eigenvectors from these equations? This time i would actually like someone to show and explain the reasoning behind the missing step from these equations to the set of eigenvectors. Don't get me wrong, i can setup the system of equations, but from there on i'm confused on how to determine the eigenvectors in terms of real numbers and not in terms of variables.

• For any given eigenvalue, there not "an" eigenvector; but an eigenspace. How would you want to express a subspace "in terms of real numbers and not in terms of variables"? – Martin Argerami Mar 17 '16 at 6:51
• Well atleast in one of the examples in my book they get three eigenvectors after some calculation on the form $\textbf{v} = \begin{pmatrix}1/2(z -y)&y&z\end{pmatrix}$ and $\textbf{v}_3 = \begin{pmatrix}-z&0&z\end{pmatrix}$ and from that deduce that the eigenvectors are: $\textbf{v}_1 = \begin{pmatrix}-1/2&1&0\end{pmatrix}, \textbf{v}_2 = \begin{pmatrix}1/2&0&1\end{pmatrix}, \textbf{v}_3 = \begin{pmatrix}-1&0&1\end{pmatrix}$. And i would like to see the same chain of reasoning from where i am to that. – Thomas Mar 17 '16 at 7:14

Your eigenvalues are correct. Let's step through one eigenvector example and you can do the other two. We will use $\lambda = 0$. Note that there are other approaches to finding eigenvectors, this is just one approach.

To find the eigenvectors, we want to find $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$.

For $\lambda_1 = 0$, $(A- \lambda_1 I)v_1 = \begin{pmatrix}2 & 1 & -3\\4 & 2 & 3\\0 & 0 & 1\end{pmatrix} v_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

Row reduction yields

$$\begin{pmatrix}1 & \dfrac{1}{2} & 0\\0 &0 & 1\\0 & 0 & 0\end{pmatrix} v_1 = \begin{pmatrix}1 & \dfrac{1}{2} & 0\\0 &0 & 1\\0 & 0 & 0\end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

In other words, for the eigenvector $v_1 = (a, b, c)$, \begin{align*} a &= -\dfrac{1}{2} b \\ c &= 0 \end{align*}

This gives us a free variable, $b$, that we can set to anything we'd like (in other words, eigenvectors are not unique). To make the eigenvector easy to write, let's choose $b = 2$, giving us the eigenvector $$v_1 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$$

Next, try finding the eigenvectors for $\lambda_2 = 1$ and $\lambda_3 = 4$.

One solution for these eigenvectors can be:

$$v_2 = \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}, v_3 \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$$