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.

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.

The post An Example of a Real Matrix that Does Not Have Real Eigenvalues first appeared on Problems in Mathematics.

Let $n$ be an odd integer and let $A$ be an $n\times n$ real matrix.
Prove that the matrix $A$ has at least one real eigenvalue.

 

We give two proofs.

Proof 1.

Let $p(t)=\det(A-tI)$ be the characteristic polynomial of the matrix $A$.
It is a degree $n$ polynomial and the coefficients are real numbers since $A$ is a real matrix. Since $n$ is odd, the leading term of $p(t)$ is $-t^n$. That is, we have
\[p(t)=-t^n+\text{lower terms}.\] (Note: if you use the alternative definition of characteristic polynomial $p(t)=\det(tI-A)$, then the leading term is $t^n$. You can easily modify the proof for this case.)

Therefore, as $t$ increases the polynomial $p(t)$ tends to $-\infty$:
\[\lim_{t\to \infty}=-\infty.\] Similarly, we have
\[\lim_{t \to -\infty}=\infty.\] By the intermediate value theorem, there is $-\infty < \lambda < \infty$ such that$p(\lambda)=0$. Since a root of $p(t)$ is an eigenvalue of $A$, we have obtained a real eigenvalue $\lambda$ of $A$.

Proof 2.

Let $p(t)=\det(A-tI)$ be the characteristic polynomial of the matrix $A$ and write it as
\[p(t)=\prod_{i=1}^k(\lambda_i-t)^{n_i},\] where $\lambda_i$ are distinct eigenvalues of $A$ and $n_i$ is the algebraic multiplicity of $\lambda_i$.

Since $A$ is a real matrix, the coefficients of $p(t)$ are real numbers. So we have
\[\overline{p(t)}=p(\,\bar{t}\,).\] Thus, we have
\begin{align*}
p(\bar{t})&=\overline{p(t)}=\prod_{i=1}^k(\,\bar{\lambda}_i-\bar{t}\,)^{n_i}.
\end{align*}
Replacing $\bar{t}$ by $t$, we obtain
\begin{align*}
p(t)=\prod_{i=1}^k(\bar{\lambda}_i-t)^{n_i}.
\end{align*}
This yields that the complex conjugate $\bar{\lambda}_i$ is an eigenvalue with algebraic multiplicity $n_i$.
Hence each non-real eigenvalue $\lambda$ appears as a pair $(\lambda, \bar{\lambda})$ counting multiplicity.
Thus, there are even number of non-real eigenvalues of $A$.

The number of all eigenvalues of $A$ counting multiplicities is $n$.
Since $n$ is odd, there must be at least one real eigenvalue of $A$.

The post There is at Least One Real Eigenvalue of an Odd Real Matrix first appeared on Problems in Mathematics.

Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\bar{A}^{\trans}$.
Prove that every Hermitian matrix $A$ can be written as the sum
\[A=B+iC,\] where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix.

Proof.

Since $A$ is Hermitian, we have
\[\bar{A}^{\trans}=A.\] Taking the conjugate of this identity, we also have
\[A^{\trans}=\bar{A}. \tag{*}\]

Let
\[B=\frac{1}{2}(A+\bar{A})\] and
\[C=\frac{1}{2i}(A-\bar{A}).\] We claim that $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix.

We have
\begin{align*}
\bar{B}=\frac{1}{2}\overline{(A+\bar{A})}=\frac{1}{2}(\bar{A}+\bar{\bar{A}})=\frac{1}{2}(\bar{A}+A)=B.
\end{align*}
Thus, the matrix $B$ is real. To prove $B$ is symmetric, we compute
\begin{align*}
&B^{\trans}=\frac{1}{2}(A+\bar{A})^{\trans}\\
&=\frac{1}{2}(A^{\trans}+\bar{A}^{\trans})\\
&=\frac{1}{2}(\bar{A}+A) && \text{by (*) and $A$ is Hermitian}\\
&=B.
\end{align*}
This proves that $B$ is symmetric.

The matrix $C$ is real because we have
\begin{align*}
\bar{C}=\frac{1}{-2i}(\bar{A}-\bar{\bar{A}})=\frac{1}{2i}(A-\bar{A})=C.
\end{align*}
We also have
\begin{align*}
&C^{\trans}=\frac{1}{2i}(A^{\trans}-\bar{A}^{\trans})\\
&=\frac{1}{2i}(\bar{A}-A)&& \text{by (*) and $A$ is Hermitian}\\
&=-\frac{1}{2i}(A-\bar{A})=-C.
\end{align*}
Hence $C$ is a skew-symmetric matrix.

Finally, we compute
\begin{align*}
B+iC&=\frac{1}{2}(A+\bar{A})+i\cdot \frac{1}{2i}(A-\bar{A})\\
&=\frac{1}{2}(A+\bar{A})+\frac{1}{2}(A-\bar{A})\\
&=A.
\end{align*}
Therefore, we have obtained the sum as described in the problem.

Related Question.

For a proof of this problem, see the post “Every complex matrix can be written as $A=B+iC$, where $B, C$ are Hermitian matrices“.

The post Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix first appeared on Problems in Mathematics.

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 $A$.

 

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 $A$.

The post Complex Conjugates of Eigenvalues of a Real Matrix are Eigenvalues first appeared on Problems in Mathematics.