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 $A$.

 
LoadingAdd to solve later

We give two proofs.

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 $A$.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
Sequence Converges to the Largest Eigenvalue of a Matrix

Let $A$ be an $n\times n$ matrix. Suppose that $A$ has real eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ with corresponding eigenvectors...

Close