Let \(A,B\) be real symmetric \(n\times n\) matrices, with \(B\) positive definite. For \(x\ne0\), define \[ G(x)=\frac{\langle Ax,x\rangle}{\langle Bx,x\rangle}. \]
- Show that \(G\) attains its maximum value.
- Show that any maximum point is an eigenvector of a matrix naturally associated to \(A\) and \(B\), and identify that matrix.
Solution
1. The function \(G\) is continuous on \(\RR^n\setminus\{0\}\) and satisfies\[ G(cx)=G(x) \]for every \(x\neq0\) and every nonzero real scalar \(c\).
Since \(B\) is positive definite, \[ \inner{Bx}{x}>0 \] for every \(x\neq0\). Thus the denominator never vanishes on the domain, so \(G\) is continuous there.
For \(c\neq0\), \[ \begin{aligned} G(cx) &= \frac{\inner{A(cx)}{cx}}{\inner{B(cx)}{cx}}\\ &= \frac{c^2\inner{Ax}{x}}{c^2\inner{Bx}{x}}\\ &= G(x). \end{aligned} \]
2. The function \(G\) attains a maximum on \(\RR^n\setminus\{0\}\).
Restrict \(G\) to the Euclidean unit sphere \[ S^{n-1}=\{x\in\RR^n:\norm x=1\}. \] This sphere is compact, and step 1 shows that \(G\) is continuous on it. Hence \(G\) attains a maximum there, say at \(u\in S^{n-1}\).
For any nonzero \(x\), step 1 gives \[ G(x)=G\left(\frac{x}{\norm x}\right). \] Thus the maximum on the unit sphere is also the maximum on the entire domain.
3. Let \(u\neq0\) be any maximum point and put\[ \lambda=G(u). \]Then\[ Au=\lambda Bu. \]
Fix any \(y\in\RR^n\) and consider \[ \varphi(t)=G(u+ty) \] for \(t\) near \(0\). Since \(u\) is a global maximum point, \(t=0\) is a local maximum of \(\varphi\), so \[ \varphi'(0)=0. \]
Set \[ N(t)=\inner{A(u+ty)}{u+ty}, \qquad D(t)=\inner{B(u+ty)}{u+ty}. \] Because \(A\) and \(B\) are symmetric, \[ N'(0)=2\inner{Au}{y}, \qquad D'(0)=2\inner{Bu}{y}. \] The quotient rule therefore gives \[ 0 = \varphi'(0) = \frac{ 2\inner{Au}{y}D(0) - 2N(0)\inner{Bu}{y} }{ D(0)^2 }. \] Since \[ \lambda=\frac{N(0)}{D(0)}, \] this is equivalent to \[ \inner{Au-\lambda Bu}{y}=0. \] As this holds for every \(y\in\RR^n\), \[ Au-\lambda Bu=0. \]
4. Every maximum point \(u\) is an eigenvector of\[ \boxed{B^{-1}A}, \]with eigenvalue \(G(u)\).
Positive definiteness implies that \(B\) is invertible: if \(Bx=0\), then \[ \inner{Bx}{x}=0, \] so positive definiteness forces \(x=0\).
Applying \(B^{-1}\) to the identity in step 3 gives \[ B^{-1}Au = \lambda u = G(u)u. \] Since \(u\neq0\), it is an eigenvector of \(B^{-1}A\).
5. Q.E.D.
Step 2 proves existence of a maximum, and step 4 identifies the required matrix and eigenvector relation.