Linear Algebra

Complex Conjugates of Eigenvalues of a Real Matrix are Eigenvalues

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 404

Let $A$ be an $n\times n$ real matrix.

Prove that if $\lambda$ is an eigenvalue of $A$, then its complex conjugate $\bar{\lambda}$ is also an eigenvalue of $A$.

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Proof 1.

Let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. Then we have
\[A\mathbf{x}=\lambda \mathbf{x}.\] Taking the conjugate of both sides, we have
\[\overline{A\mathbf{x}}=\overline{\lambda \mathbf{x}}.\]

Since $A$ is a real matrix, it yields that
\[A\bar{\mathbf{x}}=\bar{\lambda}\bar{\mathbf{x}}. \tag{*}\] Note that $\mathbf{x}$ is a nonzero vector as it is an eigenvector. Then the complex conjugate $\bar{\mathbf{x}}$ is a nonzero vector as well.
Thus the equality (*) implies that the complex conjugate $\bar{\lambda}$ is an eigenvalue of $A$ with corresponding eigenvector $\bar{\mathbf{x}}$.

Proof 2.

Let $p(t)$ be the characteristic polynomial of $A$.
Recall that the roots of the characteristic polynomial $p(t)$ are the eigenvalues of $A$.
Thus, we have
\[p(\lambda)=0.\]

As $A$ is a real matrix, the characteristic polynomial $p(t)$ has real coefficients.
It follows that
\[\overline{p(t)}=p(\,\bar{t}\,).\] The previous two identities yield that
\begin{align*}
p(\bar{\lambda})=\overline{p(\lambda)}=\bar{0}=0,
\end{align*}
and the complex conjugate $\bar{\lambda}$ is a root of $p(t)$, and hence $\bar{\lambda}$ is an eigenvalue of $A$.

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