Prove that the eigenvalues of a real symmetric matrix are real

Question

I am having a difficult time with the following question. Any help will be much appreciated.

Let $A$ be an $n×n$ real matrix such that $A^T = A$. We call such matrices “symmetric.” Prove that the eigenvalues of a real symmetric matrix are real (i.e. if $\lambda$ is an eigenvalue of $A$, show that $\lambda = \overline{\lambda}$ )

Answer 1

Hint: Prove that $$x^\ast A x=\langle x , A x\rangle = \langle Ax, x\rangle = x^\ast A^\ast x $$ Where $A^\ast=\overline{A}^T$

Answer 2

Let $(\lambda,v)$ be any eigenpair of $A$. Since $A=A^T=A^\ast$, $$\langle Av,Av\rangle=v^*A^*Av=v^\ast A^2v=v^*(A^2v)=\lambda^2||v||^2.$$

Therefore $\lambda^2=\frac{\langle Av,Av\rangle}{||v||^2}$ is a real nonnegative number. Hence $\lambda$ must be real.

Answer 3

Hint: for every $n\times n$ matrix $M$ $$ \langle Mv , w\rangle ~=~ \langle v , M^H w\rangle $$ where $M^H$ is the conjugate transpose of $M$ and $\langle\,\cdot\,,\,\cdot\,\rangle$ is the complex inner product (i.e. $\langle v,w\rangle=v^Hw$).

1. Think about how the eigenvalues of $M^H$ and those of $M$ are related
2. Let $v$ be a $\lambda$-eigenvector of $M$ and try what happens choosing $v=w$ in the above equation
3. Now, if $A$ is real valued and symmetric then $A^H=A$; try again the second point with $M=A$...

Answer 4

We are given that A is real symmetric, i.e \begin{align*} A = A^T \end{align*} If A were to have complex eigenvalues, then we can write \begin{align*} Ax = \lambda x \\ A\bar{x} = \bar{\lambda}\bar{x} \end{align*} Under complex conjugation, we can write \begin{align} \bar{x}^TAx = \bar{x}^T\lambda x = \lambda ||x||^2 \tag{i} \\ x^TA\bar{x} = x^T\bar{\lambda}x = \bar{\lambda}||x||^2 \tag{ii}.\\ \end{align} Since A is symmetric, $$\begin{align} \bar{x}^TAx = & (Ax)^{T} \bar{x} \\ = & x^{T} A^{T} \bar{x} \\ = & x^{T} A \bar{x}. \end{align}$$ Subtracting (i) from (ii), we get \begin{align*} \bar{\lambda}||x||^2 - \lambda ||x||^2 = 0\\ (\bar{\lambda}-\lambda)||x||^2 = 0 \end{align*} Only way this is possible for a non-zero z is if \begin{align*} \lambda = \bar{\lambda} \end{align*} Therefore, $\lambda$ is real.

Answer 5

Consider the real operator $$u := (x \mapsto Ax)$$ for all $x \in \mathbb{R}^{n}$ and the complex operator $$\tilde{u} := (x \mapsto Ax) $$ for all $x \in \mathbb{C}^{n}$. Both operators have the same characteristic polynomial, say $p(\lambda) = \det(A - \lambda I)$. Since $A$ is symmetric, $\tilde{u}$ is an hermitian operator. For the spectral theorem for hermitian operators all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $\tilde{u}$ are real. Hence, all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $u$ are real.

We have shown that the eigenvalues of a symmetric matrix are real numbers as a consequence of the fact that the eigenvalues of an Hermitian matrix are reals.