# Proof review: Symmetric matrices have real eigenvalues

This document provides the following proof:

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 $$\mathbf{v} \cdot \mathbf{v} \ge 0$$ 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 $$\lambda \in \mathbb{R}$$.

How does the author get 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 $$\overline{\lambda}(\overline{\mathbf{v}} \cdot \mathbf{v})$$?

I would appreciate it if someone could please take the time to clarify this.

• $\lambda$ is just a scalar. For example, the dot product of $x$ and $2y$ is just $2$ times the dot product of $x$ and $y$. Commented Sep 26, 2019 at 22:37
• Try writing $\mathbf{v} = \begin{bmatrix} v_1 & v_2 & \cdots & v_n\end{bmatrix}^T$; the result will fall out quite simply. Commented Sep 26, 2019 at 22:42

Apparently, the author defines $$x\cdot y=x^Ty$$, even when $$x$$ or $$y$$ are complex vectors. This is a bit different from the definition of the usual inner product $$\langle x,y\rangle=\overline{y}^Tx$$ (or $$\langle x,y\rangle=\overline{x}^Ty$$, depending on convention).

Thus $$\overline{v}^T(\lambda v)=\lambda(\overline{v}^Tv)=\lambda(\overline{v}\cdot v)$$ and $$(\overline{\lambda}\overline{v}^T)v=\overline{\lambda}(\overline{v}^Tv)=\overline{\lambda}(\overline{v}\cdot v)$$.

The dot product can be indicated by

$$\vec w \cdot \vec v$$

or equivalently

$$\vec w^T\vec v$$

or also

$$\langle \vec w,\vec v\rangle$$

and we can move the scalar factor $$\lambda$$ in any position, that is

$$\lambda\vec w \cdot \vec v=\vec w \cdot \lambda\vec v=\langle \lambda\vec w,\vec v\rangle=\cdots$$