Roots of the characteristic polynomial of a symmetric matrix I'm looking for a proof (using basic tools : definition of the characteristic polynomial and its basic properties) of the following fact : 
The roots of the characteristic polynomial of a symmetric matrix (with real coefficients) are reals.

Thank you.
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Roots are the eigenvalues. So, we consider the eigenvalues equation of $M = M\+$. If an eigenvalue is zero, it is proved. So, we consider non null eigenvalues $\braces{\lambda}$:
$$
\left.%
\begin{array}{rcl}
M{\bf v} &amp; = &amp; \lambda{\bf v}\,,\
\mbox{with}\ {\bf v}\ \mbox{non null}
\\
{\bf v}\+\lambda^{*} &amp; = &amp; {\bf v}\+M\+
\end{array}\right\rbrace
\quad\imp\quad
0 = \pars{\lambda - \lambda^{*}}{\bf v}\+ M{\bf v}
=
\pars{\lambda - \lambda^{*}}\lambda\,{\bf v}\+{\bf v}
$$
Then, $\quad\color{#0000ff}{\large \lambda^{*} = \lambda}$.
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 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
Roots are the eigenvalues. So, we consider the eigenvalues equation of $M = M\+$. If an eigenvalue is zero, it is proved. So, we consider non null eigenvalues $\braces{\lambda}$:
$$
\left.%
\begin{array}{rcl}
M{\bf v} &amp; = &amp; \lambda{\bf v}\,,\
\mbox{with}\ {\bf v}\ \mbox{non null}
\\
{\bf v}\+\lambda^{*} &amp; = &amp; {\bf v}\+M\+
\end{array}\right\rbrace
\quad\imp\quad
0 = \pars{\lambda - \lambda^{*}}{\bf v}\+ M{\bf v}
=
\pars{\lambda - \lambda^{*}}\lambda\,{\bf v}\+{\bf v}
$$
Then, $\quad\color{#0000ff}{\large \lambda^{*} = \lambda}$.