We are given a $3 \times 3$ real matrix $A$, and we know it has three eigenvalues. One eigenvalue is $\lambda_1=-1$ with corresponding eigenvector $v_1=\left[\begin{matrix}
0 \\
1 \\
0 \\
\end{matrix}\right]$ and another eigenvalue $\lambda_2=1+i$ and corresponding eigenvector $v_2=\left[\begin{matrix}
1 \\
2 \\
i \\
\end{matrix}\right]$. Given this, how can we find the third eigenvalue/eigenvector pair $(\lambda_3, v_3)$? The point is ultimately to be able to find the general solution to the linear DE system $x'=Ax$.
The context is that this problem came up in a qualifying exam. My linear algebra is incredibly rusty so I imagine there's just some eigenvalue/eigenvector related trick I'm not seeing. Now, considering the characteristic polynomial, it should be clear that the third eigenvalue is $\lambda_3 = 1-i$. What's not clear to me is determining the corresponding eigenvector. Clearly, it must be linearly independent from the other two, but how can we use the given eigenvectors to deduce the third one?
Best Answer
Hint:
The characteristic polynomial $\chi_A$ has real coefficients, hence its nonreal roots are pairwise conjugate, so $1-i$ is the third eigenvalue. Furthermore, if $v_2$ is an eigenvector associated to the eigenvalue $1+i$, $\bar v_2$ is an eigenvector associated to the conjugate eigenvalue $1-i$.