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

Under ground
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  • Every real symmetric matrix $A$ has at least one eigenvalue $\lambda$. To prove that it is real, see here:http://math.stackexchange.com/questions/1209094/proving-the-eigenvalues-of-a-real-symmetric-matrix-are-real?lq=1 – Surb Apr 02 '15 at 06:54
  • The $0\times 0$ matrix does not have any (real or other) eigenvalues. – Marc van Leeuwen Apr 02 '15 at 07:42

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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.

Marc van Leeuwen
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  • I might be misunderstanding what you mean by "tangent direction", but I don't see why $Q$ has zero derivative in every tangent direction at $v_m$. For example if $Q(x,y)=x^2+y^2$ (so $A$ is the $2\times 2$ identity matrix), then $Q$ is constant along the unit circle, and it's not true that every directional derivative of $Q$ is zero along the unit circle. – Ehsaan Nov 25 '21 at 21:28
  • @Ehsaan: I don't get what you want to say. Every directional derivative of $Q$ at a point of the unit circle in the direction tangent to the unit circle at that point is zero. Simply because one can compute a directional derivative at $p$ in direction $v$ along any parametrised curve $\gamma$ with $\gamma(0)=p$ and $\gamma'(0)=v$; in particular one can arrange $\gamma$ to have its images contained in the unit circle, in which case $Q(\gamma(t))$ is constant and its derivative $0$. – Marc van Leeuwen Nov 26 '21 at 13:06