Minimum of a quadratic form

Question

If $\bf{A}$ is a real symmetric matrix, we know that it has orthogonal eigenvectors. Now, say we want to find a unit vector $\bf{n}$ that minimizes the form: $${\bf{n}}^T{\bf{A}}\ {\bf{n}}$$ How can one prove that this vector is given by the eigenvector corresponding to the minimum eigenvalue of $\bf{A}$?

I have a proof of my own, but it's rather unelegant - how would you go about proving this?

Answer 1

Let $A=UDU^T$ be its eigen decomposition. Then $D$ is a diagonal matrix with all the eigenvalues as diagonal entries. Then we have \begin{align} \min_{n^Tn=1}~n^TAn\\&=\min_{u^Tu=1}u^TDu ~~~\{u=Un\} \\ &=\min_{x\in\mathbb{S}}\sum_{i}x_iD_{ii}~~~~~~~~~ \end{align} where $$\mathbb{S}=\{(x_1,\dots,x_N)\in\mathbb{R}^N\mid x_i\geq 0,~~\sum_{i}x_i=1\}$$ The last step is equivalent to \begin{align} \min_{x\in\mathbb{S}}\sum_{i}x_iD_{ii}=\min_{i}D_{ii} \end{align} and also note that \begin{align} \max_{x\in\mathbb{S}}\sum_{i}x_iD_{ii}=\max_{i}D_{ii} \end{align}

Answer 2

For a symmetric matrix A, there exists a diagonal matrix D and an orthonormal matrix S such that $S^{-1}AS=D $ where the diagonal entries of D are the eigenvalues of A and the rows of S are the eigenvectors of A. so if we let $n $ be the eigenvector corresponding to the smallest eigenvalue of A then $n^TAn=\lambda $

To push this proof further;

A unit eigenvector $v$ with corresponding eigenvalue $\lambda $ satisfies;

$Av=\lambda v \Rightarrow v^TAv=\lambda$