every real symmetric matrix has at least one real eigenvalue.

Question

Why every real symmetric matrix has at least one real eigenvalue?

Answer 1

Since none of the linked questions really address the question of existence (as opposed to realness) of eigenvalues at all (the complex numbers probably being assumed as workhorse here), I'll suggest an approach that avoids depending on complexification.

Let $V=\Bbb R^n$ with $n>0$ (a necessary assumption, see my comment) equipped with its standard inner product. Then let $B(v,w)=( v\mid A(w))$ be the symmetric bilinear form on $V$ defined by$~A$, and $Q:v\mapsto B(v,v)$ its quadratic form. The latter is a differentiable real-valued function, with deriviative at $v$ in the direction $w$ given by $Q'(v)(w)=B(v,w)+B(w,v)=2B(v,w)$.

By continuity, $Q$ assumes a maximal value on the (non-empty and compact) unit sphere, say at $v_m$. Then the derivative of $Q$ at $v_m$ in any tangent direction to the unit sphere at$~v_m$ is zero. Those tangent directions are all $w\in V$ with $w\perp v_m$; that is one has $$ 2B(v_m,w) = 0 \qquad\text{whenever $w\perp v_m$.} $$ Given that $B(v,w)=( v\mid A(w)) = ( A(v)\mid w)$ one concludes that $A(v_m)$ (being perpendicular to all vectors $w\perp v_m$) is a scalar multiple of$~v_m$, so $v_m$ is an eigenvector (over$~\Bbb R$) of$~A$, QED.

N.B., you can extend this to the construction of a orthonormal basis of eigenvectors for$~A$ by restricting $B$ and $Q$ to the orthogonal complement $\langle v_m\rangle^\perp$, and then applying induction on the dimension.