Derivative of fraction of vectors

Question

Hi is there anyone that can help me with this calculation?

So, I don’t know the exact way to differentiate the fraction of vector function. I tried to use the formal way but the components of the numerator had different size of dimension. Btw I actually solved the question by Cauchy-Schwarts inequality but am just curious with the technique the solution suggested. Thanks.

Answer 1

$ \newcommand\R{\mathbb R} $

I assume that $x'$ means the transpose of $x$.

By differentiate, I assume they mean take gradient $\nabla_\alpha$. The facts we need are

1. The product rule holds. If $f(\alpha)$ and $g(\alpha)$ are scalar-valued functions then $$ \nabla_\alpha f(\alpha)g(\alpha) = (\nabla_\alpha f(\alpha))g(\alpha) + f(\alpha)(\nabla_\alpha g(\alpha)). $$
2. If $f : \R \to \R$ and $g : \R^n \to \R$ then the chain rule is $$ \nabla_\alpha f(g(\alpha)) = \frac{\partial f}{\partial g}\nabla g(\alpha). $$
3. If $x$ is a vector independent of $\alpha$ then $$ \nabla_\alpha(\alpha'x) = \nabla_\alpha(x'\alpha) = x. $$

All together, we see $$\begin{aligned} \nabla_\alpha\rho_{yw}^2 &= \frac{2(\alpha'\sigma_{yx})[\nabla_\alpha(\alpha'\sigma_{yx})]}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)} - \frac{(\alpha'\sigma_{yx})^2}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)^2}[\nabla_\alpha(\alpha'\Sigma_{xx}\alpha)] \\ &= \frac{2\alpha'\sigma_{yx}}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)}\sigma_{yx} - \frac{(\alpha'\sigma_{yx})^2}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)^2}[\nabla_\beta(\beta'\Sigma_{xx}\alpha) + \nabla_\beta(\alpha'\Sigma_{xx}\beta)] \\ &= \frac{2\alpha'\sigma_{yx}}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)}\sigma_{yx} - \frac{(\alpha'\sigma_{yx})^2}{\sigma_{yy}(\alpha'\Sigma_{xx}\alpha)^2}[\Sigma_{xx} + \Sigma_{xx}']\alpha. \end{aligned}$$ The first equality is application of the product rule followed by application of the chain rule in each term. The second equality is application of fact (3.) in the first term and application of the product rule in the second term. Finally, the third equality follows application of fact (3.) in the second term, noting that $\alpha'\Sigma_{xx}\beta = (\Sigma_{xx}'\alpha)'\beta$. I assume now that $\Sigma_{xx}$ is symmetric so that $\Sigma_{xx} + \Sigma_{xx}' = 2\Sigma_{xx}$. Setting $\nabla_\alpha\rho_{yw}^2 = 0$ and making the obvious cancellations gives $$ \sigma_{yx} = \frac{\alpha'\sigma_{yx}}{\alpha'\Sigma_{xx}\alpha}\Sigma_{xx}\alpha \implies \alpha = \frac{\alpha'\Sigma_{xx}\alpha}{\alpha'\sigma_{yx}}\Sigma_{xx}^{-1}\sigma_{yx}. $$