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

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 2

Let $Ax=\lambda x$ with $x\ne 0$, with $\lambda\in\mathbb{C}$, then \begin{align} \lambda \bar x^T x &= \bar x^T(\lambda x)\\ &=\bar x^T A x \\ &=(A^T \bar{x})^T x \\ &=(A \bar x)^T x \\ &=(\bar A \bar x)^T x \\ &=(\bar\lambda\bar x)^T x\\ &=\bar \lambda \bar x^T x.\\ \end{align} Because $x\ne 0$, then $\bar{x}^T x\ne 0$ and $\lambda=\bar \lambda$.

Answer 3

If $\lambda$ is any eigenvalue of a Hermitian (in particular real symmetric) matrix $A$, then for some non-zero vector $x$, $$Ax = \lambda x \implies x^*Ax = \lambda x^*x \implies \lambda = \dfrac{x^*Ax}{x^*x}.$$

Now $$\lambda^* = \dfrac{x^* A^* x}{x^*x} = \dfrac{x^*Ax}{x^*x} = \lambda.$$ Therefore, $\lambda$ is real.

Note: $(AB)^* = B^*A^*$.

Answer 4

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 5

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 $x$ is if \begin{align*} \lambda = \bar{\lambda} \end{align*} Therefore, $\lambda$ is real.

Answer 6

I found the proof in this document to be informative and educational.

The Spectral Theorem states that if $A$ is an $n \times n$ symmetric matrix with real entries, then it has $n$ orthogonal eigenvectors. The first step of the proof is to show that all the roots of the characteristic polynomial of $A$ (i.e. the eigenvalues of $A$) are real numbers.

Recall that if $z = a + bi$ is a complex number, its complex conjugate is defined by $\bar{z} = a − bi$. We have $z \bar{z} = (a + bi)(a − bi) = a^2 + b^2$, so $z\bar{z}$ is always a nonnegative real number (and equals $0$ only when $z = 0$). It is also true that if $w$, $z$ are complex numbers, then $\overline{wz} = \bar{w}\bar{z}$.

Let $\mathbf{v}$ be a vector whose entries are allowed to be complex. It is no longer true that $\mathbf{v} \cdot \mathbf{v} \ge 0$ with equality only when $\mathbf{v} = \mathbf{0}$. For example,

$$\begin{bmatrix} 1 \\ i \end{bmatrix} \cdot \begin{bmatrix} 1 \\ i \end{bmatrix} = 1 + i^2 = 0$$

However, if $\bar{\mathbf{v}}$ is the complex conjugate of $\mathbf{v}$, it is true that $\bar{\mathbf{v}} \cdot \mathbf{v} \ge 0$ with equality only when $\mathbf{v} = 0$. Indeed,

$$\begin{bmatrix} a_1 - b_1 i \\ a_2 - b_2 i \\ \dots \\ a_n - b_n i \end{bmatrix} \cdot \begin{bmatrix} a_1 + b_1 i \\ a_2 + b_2 i \\ \dots \\ a_n + b_n i \end{bmatrix} = (a_1^2 + b_1^2) + (a_2^2 + b_2^2) + \dots + (a_n^2 + b_n^2)$$

which is always nonnegative and equals zero only when all the entries $a_i$ and $b_i$ are zero.

With this in mind, suppose that $\lambda$ is a (possibly complex) eigenvalue of the real symmetric matrix $A$. Thus there is a nonzero vector $\mathbf{v}$, also with complex entries, such that $A\mathbf{v} = \lambda \mathbf{v}$. By taking the complex conjugate of both sides, and noting that $A = A$ since $A$ has real entries, we get $\overline{A\mathbf{v}} = \overline{\lambda \mathbf{v}} \Rightarrow A \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}$. Then, using that $A^T = A$,

$$\overline{\mathbf{v}}^T A \mathbf{v} = \overline{\mathbf{v}}^T(A \mathbf{v}) = \overline{\mathbf{v}}^T(\lambda \mathbf{v}) = \lambda(\overline{\mathbf{v}} \cdot \mathbf{v}),$$

$$\overline{\mathbf{v}}^T A \mathbf{v} = (A \overline{\mathbf{v}})^T \mathbf{v} = (\overline{\lambda} \overline{\mathbf{v}})^T \mathbf{v} = \overline{\lambda}(\overline{\mathbf{v}} \cdot \mathbf{v}).$$

Since $\mathbf{v} \not= \mathbf{0}$,we have $\overline{\mathbf{v}} \cdot \mathbf{v} \not= 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in \mathbb{R}$.

For further information on how the author gets from $\overline{\mathbf{v}}^T(\lambda \mathbf{v})$ to $\lambda(\overline{\mathbf{v}} \cdot \mathbf{v})$ and from $(\overline{\lambda} \overline{\mathbf{v}})^T \mathbf{v}$ to $\overline{\lambda}(\overline{\mathbf{v}} \cdot \mathbf{v})$, see this question.

Answer 7

Since $A^* = A$, $(x^*Ax)^* = x^*Ax$. Therefore $x^*Ax$ is a real number for any $x$. If $x$ is an eigenvalue of $A$ with eigenvalue $\lambda$, we have $x^*Ax = x^*(\lambda x) = \lambda x^*x$. Since $x^*Ax$ and $x^*x$ are always real (and $x^*x$ is not zero for an eigenvector $x$), this means $\lambda$ must be real too.

Answer 8

Consider an eigen-pair, $(\lambda,v)$, of an $n\times n$ real symmetric matrix $A$,

then $ \begin{align*} Av=\lambda v \implies \bar{Av}=\bar{\lambda v}=A\bar{v}=\bar{\lambda}\bar{v}. \end{align*} $

Now,

$ \begin{align*} \left\langle \bar{v},Av \right\rangle =\bar{v}^\top Av =\bar{v}\lambda v =\lambda \left\langle \bar{v}, v\right\rangle. \end{align*} $

Similarly,

$ \begin{align*} \left\langle \bar{v},Av \right\rangle =\bar{v}^\top Av=\bar{A^\top v}^\top v =\bar{\lambda v}\lambda v =\bar{\lambda} \left\langle \bar{v}, v\right\rangle. \end{align*} $

So, we have

$ \begin{align*} & \left\langle \bar{v},Av \right\rangle = \lambda \left\langle \bar{v}, v\right\rangle = \bar{\lambda} \left\langle \bar{v}, v\right\rangle , \left\langle \bar{v}, v\right\rangle \neq 0 \\ & \implies \bar{\lambda} =\lambda \implies \lambda \in \mathbb{R}. \end{align*} $.

Answer 9

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 10

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.

Answer 11

$$A^*=A\implies(\lambda x)^*x=(Ax)^*x=(\lambda x)^*x=\lambda^*(x^*x)=x^*Ax=\lambda(x^*x)$$