Pick an orthonormal ordered basis
\(B\) of
\(V\text{,}\) and let
\(A=[T]_B\text{.}\) By
Theorem 5.4.2,
\(A\) is symmetric. To prove that all roots of the characteristic polynomial
\(p(t)=\det(tI-A)\) are real, we make a slight detour into complex vector spaces. The set
\begin{equation*}
\C^n=\{(z_1,z_2,\dots, z_n)\colon z_i\in \C \text{ for all } 1\leq i\leq n\}
\end{equation*}
of all complex \(n\)-tuples, together with the operations
\begin{equation*}
(z_1,z_2,\dots, z_n)+(w_1, w_2,\dots, w_n)=(z_1+w_1, z_2+w_2, \dots, z_n+w_n)
\end{equation*}
and
\begin{equation*}
\alpha (z_1, z_2, \dots, z_n)=(\alpha z_1, \alpha z_2, \dots, \alpha z_n),
\end{equation*}
where
\(\alpha\in \C\text{,}\) forms what is called a
vector space over \(\C\). This means that
\(V=\C^n\) satisfies the strengthened axioms of
Definition 3.1.1 obtained by replacing every mention of a scalar
\(c\in \R\) with a scalar
\(\alpha\in \C\text{.}\) Additionally, the vector space
\(\C^n\) has the structure of a complex inner product defined as
\begin{equation*}
\langle (z_1,z_2,\dots, z_n), (w_1,w_2,\dots, w_n)\rangle=z_1\overline{w_1}+z_2\overline{w_2}+\cdots +z_n\overline{w_n}\text{,}
\end{equation*}
where \(\overline{w_i}\) denotes the complex conjugate of \(w_i\) for each \(i\text{.}\) Essentially all of our theory of real vector spaces can be “transported” to complex vector spaces, including definitions and results about eigenvectors and inner products. The rest of this argument makes use of this principle by citing without proof some of these properties, and this is why it has been downgraded to a “proof sketch”.
We now return to \(A\) and its characteristic polynomial \(p(x)\text{.}\) Recall that we want to show that all roots of \(p(x)\) are real. Let \(\lambda\in \C\) be a root of \(p(x)\text{.}\) The complex theory of eigenvectors implies that there is a nonzero vector \(\boldz\in \C^n\) satisfying \(A\boldz=\lambda \boldz\text{.}\) On the one hand, we have
\begin{equation*}
\langle A\boldz, \boldz\rangle =\langle \lambda\boldz, \boldz\rangle=\lambda\langle \boldz, \boldz\rangle
\end{equation*}
using properties of our complex inner product. On the other hand, since
\(A^T=A\) it is easy to see that
Corollary 5.4.3 extends to our complex inner product: i.e.,
\begin{equation*}
\langle A\boldz, \boldw\rangle=\langle \boldz, A\boldw\rangle
\end{equation*}
for all \(\boldz, \boldw\in \C^n\text{.}\) Thus
\begin{align*}
\langle A\boldz, \boldz\rangle \amp= \\
\amp= \langle \boldz, A\boldz\rangle \\
\amp = \langle \boldz, \lambda\boldz\rangle\\
\amp =\overline{\lambda}\langle \boldz, \boldz\rangle \text{.}
\end{align*}
(In the last equality we use the fact that our complex inner product satisfies \(\langle \boldz, \alpha\boldw\rangle=\overline{\alpha} \langle \boldz, \boldw\rangle\) for any \(\alpha\in \C\) and vectors \(\boldz, \boldw\in \C^n\text{.}\)) It follows that
\begin{equation*}
\lambda\langle \boldz, \boldz\rangle=\overline{\lambda}\langle \boldz, \boldz\rangle\text{.}
\end{equation*}
Since \(\boldz\ne \boldzero\text{,}\) we have \(\langle \boldz, \boldz\rangle\ne 0\) (another property of our complex inner product), and thus \(\lambda=\overline{\lambda}\text{.}\) Since a complex number \(z=a+bi\) satisfies \(\overline{z}=z\) if and only if \(b=0\) if and only if \(z\) is real, we conclude that \(\lambda\) is a real number, as claimed.
