TheoremBase

A Maximiser of the Quadratic Form of a Symmetric Matrix on Unit Vectors Orthogonal to Given Eigenvectors Is an Eigenvector

For a real symmetric matrix and finitely many (fewer than n) orthonormal eigenvectors, the quadratic form attains a maximum on the unit vectors orthogonal to them, and every maximiser is an eigenvector whose eigenvalue is the maximum value.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let n∈Nn\in\mathbb{N} and A∈S(n)A\in\mathcal{S}(n), the set of symmetric matrices, with the matrix-vector product, the dot product and the Euclidean norm ∥⋅∥\lVert\cdot\rVert on Rn\mathbb{R}^{n}, and eigenvectors of AA. Let k∈N0k\in\mathbb{N}_{0} with k<nk<n, and let u:[k]→Rnu:[k]\to\mathbb{R}^{n} be a map such that, for all i,j∈[k]i,j\in[k], ui⋅uj=1u_{i}\cdot u_{j}=1 if i=ji=j and ui⋅uj=0u_{i}\cdot u_{j}=0 if i≠ji\neq j, and such that for every i∈[k]i\in[k] there is μ∈R\mu\in\mathbb{R} for which uiu_{i} is an eigenvector of AA with eigenvalue μ\mu. Let

K={x∈Rn:∥x∥=1 and x⋅ui=0 for every i∈[k]}.K=\{x\in\mathbb{R}^{n}:\lVert x\rVert=1\text{ and }x\cdot u_{i}=0\text{ for every }i\in[k]\}.

There is v∈Kv\in K with x⋅(Ax)≤v⋅(Av)x\cdot(Ax)\le v\cdot(Av) for every x∈Kx\in K.

If v∈Kv\in K and x⋅(Ax)≤v⋅(Av)x\cdot(Ax)\le v\cdot(Av) for every x∈Kx\in K, then vv is an eigenvector of AA with eigenvalue v⋅(Av)v\cdot(Av).

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