Eigenvalues and the Spectral Theorem

For reality, allow complex vectors momentarily and let $x^\ast=\bar x^\top$ be the conjugate transpose. If $Ax=\lambda x$ with $x\neq 0$, then $x^\ast Ax=\lambda\,x^\ast x=\lambda\norm x^2$. A real symmetric $A$ is Hermitian, $A^\ast=\bar A^\top=A^\top=A$, so $(x^\ast Ax)^\ast=x^\ast A^\ast x=x^\ast Ax$ and the scalar is real. As $\norm x^2=\sum_i\abs{x_i}^2>0$ is real, $\lambda=x^\ast Ax/\norm x^2$ is real, and a real eigenvalue yields a real eigenvector from the null space of the singular real matrix $A-\lambda I$. For orthogonality, let $Au=\lambda u$ and $Av=\mu v$ with $\lambda\neq\mu$. Then $\lambda\ip uv=\ip{Au}v=\ip u{Av}=\mu\ip uv$ by symmetry, so $(\lambda-\mu)\ip uv=0$ forces $\ip uv=0$.

Argue by induction on $n$, the case $n=1$ being trivial. The Rayleigh quotient $x\mapsto x^\top Ax$ is continuous on the unit sphere $\{\norm x=1\}$, which is closed and bounded, so by the extreme value theorem it attains a maximum $\lambda_1=q_1^\top Aq_1$ at some unit vector $q_1$. For any $v\perp q_1$ the curve $t\mapsto q_1\cos t+v\sin t$ stays on the sphere, so its Rayleigh value $\lambda_1\cos^2 t+2(q_1^\top Av)\cos t\sin t+(v^\top Av)\sin^2 t$ is maximised at $t=0$; the vanishing derivative there gives $q_1^\top Av=0$, using $A=A^\top$. Thus $Aq_1\perp v$ for every $v\perp q_1$, forcing $Aq_1\in\operatorname{span}(q_1)$, so $Aq_1=\lambda_1 q_1$ with $\lambda_1=q_1^\top Aq_1$ the attained maximum, the largest Rayleigh value. The orthogonal complement $q_1^\perp$ is invariant under $A$, since for $x\perp q_1$, $\ip{Ax}{q_1}=\ip x{Aq_1}=\lambda_1\ip x{q_ 1}=0$, so $Ax\perp q_1$. Fix an orthonormal basis $e_2,\dots,e_n$ of $q_1^\perp$; by invariance the restriction of $A$ is an operator on $q_1^\perp$ with matrix $B_{ij}=\ip{Ae_j}{e_i}$, and symmetry gives $B_{ij}=\ip{Ae_j}{e_i}=\ip{e_j}{Ae_i}=B_{ji}$, so $B$ is a genuine $(n-1)\times(n-1)$ symmetric matrix. By induction it has an orthonormal eigenbasis, which pulls back to an orthonormal eigenbasis $q_2,\dots,q_n$ of $q_1^\perp$, and together with $q_1$ these form an orthonormal eigenbasis of $\R^n$. Placing the $q_i$ as the columns of $Q$ makes $Q$ orthogonal, and $AQ=Q\Lambda$ reads off the eigen-equations, giving $A=Q\Lambda Q^\top$.

Expand a unit vector in the eigenbasis, $x=\sum_i c_i q_i$ with $\sum_i c_i^2=1$. Since $Aq_i=\lambda_i q_i$ and $\ip{q_i}{q_j}=\delta_{ij}$, $x^\top Ax=\sum_{i,j}c_ic_j\lambda_j\delta_{ij}=\sum_i \lambda_i c_i^2$, a weighted average of the eigenvalues with weights $c_i^2$ summing to $1$, which lies between $\lambda_n$ and $\lambda_1$ and reaches $\lambda_1$ at $x=q_1$ and $\lambda_n$ at $x=q_n$. For the intermediate eigenvalue, given any $k$-dimensional $V$, the span of $q_k,\dots,q_n$ has dimension $n-k+1$, so it meets $V$ in a nonzero vector $x$, on which $x^\top Ax=\sum_{i\ge k}\lambda_i c_i^2\le\lambda_k$, so the inner minimum over $V$ is at most $\lambda_k$. Taking $V=\operatorname{span}(q_1,\dots,q_k)$, every unit $x\in V$ has $x^\top Ax=\sum_{i\le k}\lambda_i c_i^2\ge\lambda_k$, so the maximum over $V$ of the inner minimum is exactly $\lambda_k$.