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Let V be a finite dimensional complex valued space and $T\in \mathcal{L}(V)$ such that the matrix of $T$ (for some basis) has only real valued entries. If $\lambda$ is an eigenvalue so is $\bar{\lambda}$.

Let $\lambda = a + b i$.

Let $v \in V$ be the corresponding eigenvector for $\lambda$. Then we know that $T(v) = \lambda v$ or: $$ \begin{align} t_{1,1} x_1 +\ldots+t_{1,n}x_n &= \lambda x_1 = (a + bi)x_1\\ t_{2,1} x_1 +\ldots+t_{2,n}x_n &= \lambda x_2 = (a + bi)x_2\\ \vdots \\ t_{n,1} x_1 +\ldots+t_{n,n}x_n &= \lambda x_n = (a + bi)x_n \end{align} $$

where $t_{i,j}$ is the entries of the above mentioned matrix, and not all $x_j$ zero.

We can re-arrange the above: $$ \begin{align} (t_{1,1}-a) x_1 +\ldots+t_{1,n}x_n &= (bi)x_1\\ t_{2,1} x_1 + (t_{2,2}-a)x_2 +\ldots+t_{2,n}x_n &= (bi)x_2\\ \vdots \\ t_{n,1} x_1 +\ldots+(t_{n,n}-a)x_n &= (bi)x_n \end{align} $$

Taking the conjugate of both sides shows that $\bar{\lambda}$ also is an eigenvalue with the corresponding eigenvector $\bar{v}=(\bar{x_1},\ldots,\bar{x_n})$.

The exercise doesn't actually task me with showing what the eigenvector would be, but it seems to be the result from my attempt, and it seems to be true according to later material (which I don't quite understand yet).

Any feedback on my above attempt would be greatly appreciated, especially easier ways to show it, or any lapses in logic.

Skurmedel
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1 Answers1

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The eigenvalue is the root of $p=det(A-xI)$ where $A$ is the matrix relativity to the basis with real values. So if $\lambda$ is a root of $p$ so is $\bar \lambda$.

Tsemo Aristide
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  • And the vector as well? Is it related to the characteristic polynomial? The book I'm following is quite famous for putting determinants last, although I am familiar with the cofactor expansion etc from other places. But the characteristic polynomial is up in the next chapter. – Skurmedel Dec 09 '16 at 19:58