Eigenvalues of a Hermitian Matrix are Real Numbers

Ohio State University exam problems and solutions in mathematics

Problem 202

Show that eigenvalues of a Hermitian matrix $A$ are real numbers.

(The Ohio State University Linear Algebra Exam Problem)
 
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We give two proofs. These two proofs are essentially the same.
The second proof is a bit simpler and concise compared to the first one.

Proof 1.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\] Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}

We also have
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x})=(A\mathbf{x})^{\trans} \bar{\mathbf{x}}=\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}.
\end{align*}
The first equality follows because the dot product $\mathbf{u}\cdot \mathbf{v}$ of vectors $\mathbf{u}, \mathbf{v}$ is commutative.

That is, we have
\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=\mathbf{v}\cdot\mathbf{u}.\] We applied this fact with $\mathbf{u}=\bar{\mathbf{x}}$ and $\mathbf{v}=A\mathbf{x}$.

Thus we obtain
\[\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}=\lambda ||\mathbf{x}||.\] Taking the complex conjugate of this equality, we have
\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda} ||\mathbf{x}||. \tag{**}\] (Note that $\bar{\bar{\mathbf{x}}}=\mathbf{x}$. Also $\overline{||\mathbf{x}||}=||\mathbf{x}||$ because $||\mathbf{x}||$ is a real number.)
Since the matrix $A$ is Hermitian, we have $\bar{A}^{\trans}=A$.

This yields that
\begin{align*}
\bar{\lambda} ||\mathbf{x}|| &\stackrel{(**)}{=} \bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=}\bar{\mathbf{x}}^{\trans} \lambda \mathbf{x}\\
&=\lambda ||\mathbf{x}||.
\end{align*}
Recall that $\mathbf{x}$ is an eigenvector, hence $\mathbf{x}$ is not the zero vector and the length $||\mathbf{x}||\neq 0$.
Therefore, we divide by the length $||\mathbf{x}||$ and get
\[\lambda=\bar{\lambda}.\]

It follows from this that the eigenvalue $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all the eigenvalues of the Hermitian matrix $A$ are real numbers.

Proof 2.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\]

Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}

Now we take the conjugate transpose of both sides and get
\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda}||\mathbf{x}||. \tag{***}\] Since $A$ is a Hermitian matrix, we have $\bar{A}^{\trans}=A$.

Then the left hand side becomes
\begin{align*}
\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=} \bar{\mathbf{x}}^{\trans}\lambda\mathbf{x}\\
&=\lambda ||\mathbf{x}||. \tag{****}
\end{align*}

Therefore comparing (***) and (****) we obtain
\[\lambda ||\mathbf{x}||=\bar{\lambda}||\mathbf{x}||.\] Since $\mathbf{x}$ is an eigenvector, it is not the zero vector and the length $||\mathbf{x}||\neq 0$.
Dividing by the length $||\mathbf{x}||$, we obtain $\lambda=\bar{\lambda}$ and this implies that $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that every eigenvalue of the Hermitian matrix $A$ is a real number.

Corollary

Every real symmetric matrix is Hermitian. Thus, as a corollary of the problem we obtain the following fact:

Eigenvalues of a real symmetric matrix are real.

Related Question.

Problem. Let $A$ be an $n\times n$ real symmetric matrix.
Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.\]

Note that the inequality makes sense because eigenvalues of $A$ are real by Corollary.

For a proof of this problem, see the post “Inequality about Eigenvalue of a Real Symmetric Matrix“.

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4 Responses

  1. 04/30/2017

    […] that the eigenvalues of a real symmetric matrix are real. (See the corollary in the post “Eigenvalues of a Hermitian matrix are real numbers“.) Let $lambda$ be a (real) eigenvalue of $A$ and let $mathbf{x}$ be a corresponding real […]

  2. 06/15/2017

    […] that the eigenvalues of a real symmetric matrices are all real numbers and it is diagonalizable by an orthogonal […]

  3. 07/19/2017

    […] The proof is given in the post Eigenvalues of a Hermitian Matrix are Real Numbers […]

  4. 07/28/2017

    […] that all the eigenvalues of a symmetric matrices are real numbers. Let $lambda_1, dots, lambda_n$ be eigenvalues of […]

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