 Solution

For the reality of the eigenvalues, one may initially allow a nonzero complex eigenvector $\mathbf{v}$ with $M\mathbf{v}=\lambda \mathbf{v}$. Since $M$ is real symmetric, it is a Hermitian matrix, andis real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf{v}$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.

For two such eigenvectors, symmetry givesDistinct eigenvalues therefore give zero dot product. Together with the prescribed unit lengths,These $n$ vectors form an orthonormal basis of ${\mathbb{R}}^n$.

Expand a real unit vector as $\mathbf{x}={\sum }_a{c}_a{\mathbf{e}}_a$. Then ${\sum }_a{c}_a^2=1$, and its Rayleigh quotient isMoreover,Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge 2$. Equality occurs precisely along the top eigendirection:

To obtain a unit vector in $S$ using just ${\mathbf{e}}_1,{\mathbf{e}}_2$, let $u=({\mathbf{e}}_1{)}_1$ and $v=({\mathbf{e}}_2{)}_1$ denote their first coordinates. If $u^2+v^2>0$, takeThen ${\alpha }_{1}u+{\alpha }_{2}v=0$ and ${\alpha }_1^2+{\alpha }_2^2=1$. Hence ${\alpha }_1{\mathbf{e}}_1+{\alpha }_2{\mathbf{e}}_2$ has first coordinate zero and unit length. If $u=v=0$, simply take ${\alpha }_1=1,{\alpha }_2=0$. In either case the Rayleigh quotient of the constructed vector isThe continuous quadratic form attains its maximum on the compact space $S$, so

For $\mathbf{x}=(0,\mathbf{y}{)}^T$, deletion of the first row and column gives ${\mathbf{x}}^{T}M\mathbf{x}={\mathbf{y}}^{T}A\mathbf{y}$, while $|\mathbf{x}|=|\mathbf{y}|$. The spectral theorem for real symmetric matrices applied to $A$ identifies its largest eigenvalue with this maximum:Combining the upper bound for every unit vector with the constructed lower bound provesThis is largest-eigenvalue interlacing for a principal submatrix: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.