Take \[ A := \begin{pmatrix} 2 & -2 \\ -2 & 5 \end{pmatrix}. \] Find an orthogonal matrix \(P\) for which \(P^{-1}AP\) is diagonal. Then find the minimum value of the dot products \(Ax \cdot x\) as \(x\) ranges through the unit vectors in \(\mathbb{R}^2\).
Solution
If \(A\) has two distinct eigenvalues, we will have \(A = PD\inverseof{P}\) where \(P\) is the matrix of eigenvectors and \(D\) has eigenvalues on the diagonal. We can compute the characteristic polynomial \[ p_\chi(x) = x^2 - (\Tr A)x + \det A = x^2 - 7x + 6 = (x-6)(x-1), \]
and so \(\spec(A) = \theset{6,1}\). Computing the kernel of \(A-\lambda I\) for each of these yields \[ \vector v_1 = \left[ \begin{array} { c } { 1 } \\ { - 2 } \end{array} \right], \vector v_2 = \left[ \begin{array} { c } { 2 } \\ { 1 } \end{array} \right], \]
And so we can write and check \[ P = \left[ \begin{array} { c c } { 1 } & { 2 } \\ { - 2 } & { 1 } \end{array} \right] \\ D = \left(\begin{array}{rr} 6 & 0 \\ 0 & 1 \end{array}\right) \\ \]
We can compute \(PP^T = \mathrm{diag}(5,5)\), so \(P\) can be made orthogonal by replacing \(P\) with \((1/\sqrt 5) P\). With this replacement, a quick computation shows that \(PDP^T = A\).
We will use the fact that \(A = PD\inverseof{P}\) where since \(A\) is symmetric and \(P\) is orthogonal. We can write \[ \inner{A\vector x}{\vector x} = \vector x^T A^T \vector x = \vector x^T (PDP^T)^T \vector x = (P^T \vector x)^T D (P^T\vector x) = \vector y^T D \vector y = \inner{\vector y}{ D\vector y}, \] where \(\vector x\in S^2 \implies P^T \vector x \definedas \vector y \in S^2\) since \(P^T\) is both orthogonal and full-rank (and thus a bijection \(S^2 \selfmap\)).
We can now expand \[ \inner{\vector y}{D \vector y} = \sum_{i=1}^2 y_i \lambda_i y_i = \sum_{i=1}^2 \lambda_i y_i^2 \]
We now note that we can take \(\vector y = \thevector{0, 1}\), in which case \(D\vector y = \thevector{0, \lambda_2} = \thevector{0, 1}\) and thus \(\inner{\vector y}{D \vector y} = 1\) is a candidate minimum.
We can write this as the constrained optimization problem \[ \text{Minimize } f(y_1, y_2) = 6y_1^2 + 1 y_2^2\\ \text{subject to } g(y_1, y_2) = y_1^2 + y_2^2 = 1 \]
where we note that this constraint is equivalent to the original \(\norm{\vector y} = \sqrt{y_1^2 + y_2^2} = 1\).
This can be approached with Lagrange multipliers, i.e. looking at where \(\nabla f = \lambda \nabla g\). This yields \[ \thevector{12y_1, 2y_2} = \lambda \thevector{2y_1, 2y_2} \implies \\ 6y_1 = \lambda y_1, ~ y_2 = \lambda y_2. \]
The second condition forces \(y_2 \in \theset{0,1}\), and solving for \(\lambda\) in this expression yields \(\lambda = 1\) and so the first condition forces \(y_1 = 0\). This leaves only one possibility, \(\vector y = \thevector{0, 1}\), which is indeed the candidate from above. Thus the minimum value is 1. \(\qed\)