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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?

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    $\begingroup$ Shouldn't complex eigenvectors also be complex conjugates, just like the eigenvalues? $\endgroup$
    – Moo
    Nov 13 '19 at 23:31
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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 $1-i$.

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You know that $Av_2=\lambda_2v_2$. Take the complex conjugate of this equation and note that $A^*=A$ since $A$ is real.

Then $Av_2^*=\lambda_2^*v_2^*$. Hence the eigenvector you require is $v_2^*$

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  • $\begingroup$ +1 Dolan a $v*$ is missing in last equation. $\endgroup$
    – MtGlasser
    Nov 14 '19 at 0:00
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    $\begingroup$ Quite right- thanks. $\endgroup$
    – S. Dolan
    Nov 14 '19 at 0:12
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$$A {v}_1 = (1 + i) {v}_1 \Rightarrow \overline{A} \overline{{v}_1} = \overline{(1 + i)} \overline{{v}_1} \iff A \overline{{v}_1} = (1 - i) \overline{{v}_1}.$$

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Sum of the Eigen values is equal to the trace(sum of the diagonal elements) of the matrix A. Since you are aware of v1 and v2, you can easily find the third Eigen value.

For the third Eigenvector,$Av_3=\lambda_3v_3$ You are aware of A, $\lambda_3$, find $v_3$.

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