# The Eigenvalues of a Real Symmetric Matrix Are Real

Let $A$ be an $n \times n$ matrix over $\mathbb{F}$. Then $\lambda \in \mathbb{F}$ is an *eigenvalue* of $A$ if there exists a non-zero vector $\mathbf{v} \in \mathbb{F}^n$ such that $A \mathbf{v}=\lambda \mathbf{v}$. Any such vector $\mathbf{v}$ is an *eigenvector* of $A$ associated with the eigenvalue $\lambda$.

Let $A$ be an $n \times n$ real symmetric matrix (i.e. entries of $A$ are real and $A^{\top} = A$). Then the eigenvalues $A$ are real (i.e. if $\lambda$ is an eigenvalue of $A$, then $\lambda=\overline{\lambda}$).

**Proof.** Note that the underlying assumption of this proposition is that the ground field is $\mathbb{C}$.

Suppose that $\lambda \in \mathbb{C}$ is an eigenvalue of $A$ and $\mathbf{v}$, with complex entries, is an eigenvector of $A$ associated with $\lambda$ so that $A\mathbf{v} = \lambda \mathbf{v}$. Note that the existence of $\lambda$ is ensured by the fundamental theorem of algebra and $\mathbf{v}$ is non-zero by the definition of eigenvectors.

Taking the complex conjugate of both sides and observing that $\overline{A}=A$ since $A$ has real entries, we have

\[\overline{A \mathbf{v}}=\overline{\lambda \mathbf{v}} \Rightarrow A \overline{\mathbf{v}}=\overline{\lambda} \overline{\mathbf{v}}.\]Note that here we used the fact that for two complex numbers $z_1$ and $z_2$, we have $\overline{z_1z_2} = \bar{z_1}\bar{z_2}$.

Since $A^{\top}=A$, we have^{1}

Since $\mathbf{v} \neq \mathbf{0}$, we have $\langle\overline{\mathbf{v}}, \mathbf{v} \rangle \neq 0$ ^{2}. Thus, $\lambda=\overline{\lambda}$ and hence $\lambda \in \mathbb{R}$.

◼

This is due to the fact that matrix multiplication is associative, $(\lambda, \mathbf{v})$ is an eigenpair of $A$, and the definition of Eulidean inner product on $\mathbb{C}^n$, where $n$ is the dimension of the matrix $A$. ↩

This is by the

*non-degeneracy*(or*positive definiteness*) of an inner product. ↩