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Let $B>0$ be positive definite $d \times d$ matrix, and $v \in \mathbb R^d$. Prove that $$\max_x \frac{(x^Tv)^2}{x^TBx}=v^TB^{-1}v$$ Hint: Change of variable $z=B^{1/2}x$

$$\begin{split}\max_z\frac{(v^TB^{-1/2}z)^2}{z^Tz}\end{split}$$

Take the derivative

$$\begin{split}\frac{(z^Tz)2(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-(v^TB^{-\frac 1 2}z)^2(2z)}{(z^Tz)^2} &=\frac{2(z^Tz)(v^TB^{-\frac 1 2}z)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}z)(v^TB^{-\frac 1 2}z)z}{z^Tzz^Tz}\\ &=\frac{2(z^T\not{z)(z^T}B^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-\frac 1 2}\not{z)(z^T}B^{-\frac 1 2}v)z}{z^T\not{zz^T}z}\\ &=\frac{2(z^TB^{-\frac 1 2}v)B^{-\frac 1 2}v-2(v^TB^{-1}v)z}{z^Tz}\end{split}$$

Probably the last cancellation is invalid, wasn't sure what to do, or how to proceed.

Rodrigo de Azevedo
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Vons
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    Let $v=B^{1/2}u$. Then your desired equality may be written as $$\begin{split}\max_z\frac{(u^Tz)^2}{z^Tz}\end{split}=(B^{-1/2}v)^T(B^{-1/2}v)=u^Tu.$$ So the purpose of the substitution evidently is to reduce the problem to the case $B=I_d$. That may make it easier to sort through the algebra. – Semiclassical Feb 13 '21 at 07:14
  • Look for a rank-1 matrix and keep in mind that the nonzero eigenvalue of a rank-1 matrix is its trace. – Rodrigo de Azevedo Feb 13 '21 at 11:34
  • https://math.stackexchange.com/q/1226524/321264 – StubbornAtom Nov 11 '21 at 20:41

1 Answers1

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Following Semiclassical's comment, it suffices to show $\max_z \frac{(u^\top z)^2}{z^\top z} = u^\top u$. The Cauchy-Schwarz inequality implies $\max_z \frac{(u^\top z)^2}{z^\top z} \le u^\top u$. To show equality is attained, take $z=u$ (or $z=cu$ for any scalar $c$).

angryavian
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