An Example of a Real Matrix that Does Not Have Real Eigenvalues

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 596

Let
\[A=\begin{bmatrix}
a & b\\
-b& a
\end{bmatrix}\] be a $2\times 2$ matrix, where $a, b$ are real numbers.
Suppose that $b\neq 0$.

Prove that the matrix $A$ does not have real eigenvalues.

 
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Proof.

Let $\lambda$ be an arbitrary eigenvalue of $A$.
Then the matrix $A-\lambda I$ is singular, where $I$ is the $2\times 2$ identity matrix.
This is equivalent to having $\det(A-\lambda I)=0$.

We compute the determinant as follows.
We have
\begin{align*}
\det(A-\lambda I)&=\begin{vmatrix}
a-\lambda & b\\
-b& a-\lambda
\end{vmatrix}\\[6pt] &=(a-\lambda)^2-b(-b)\\
&=a^2-2a\lambda+\lambda^2+b^2\\
&=\lambda^2-2a\lambda+a^2+b^2.
\end{align*}

We solve the equation $\lambda^2-2a\lambda+a^2+b^2=0$ by the quadratic formula and obtain
\begin{align*}
\lambda &=\frac{2a\pm\sqrt{4a^2-4(a^2+b^2)}}{2}=\frac{2a\pm\sqrt{-4b^2}}{2}\\[6pt] &=a\pm |b|i.
\end{align*}

Since $b\neq 0$ by assumption, the eigenvalue $\lambda=a\pm|b|i$ is not a real number.
As $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all eigenvalues of $A$ are not real numbers.


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