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.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ 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 $\braces{\lambda}$: $$ \left.% \begin{array}{rcl} M{\bf v} & = & \lambda{\bf v}\,,\ \mbox{with}\ {\bf v}\ \mbox{non null} \\ {\bf v}\+\lambda^{*} & = & {\bf v}\+M\+ \end{array}\right\rbrace \quad\imp\quad 0 = \pars{\lambda - \lambda^{*}}{\bf v}\+ M{\bf v} = \pars{\lambda - \lambda^{*}}\lambda\,{\bf v}\+{\bf v} $$ Then, $\quad\color{#0000ff}{\large \lambda^{*} = \lambda}$.