Simple linear regression without intercept finding estimator for Beta in a matrix form: $\beta = \dfrac{x^{T}}{\left\|x \right\|^{2}}y$

Question

I can't think about a way how to get to this, using linear algebra. Its just simple linear regression model without intercept.

$\beta = \dfrac{x^{T}}{\left\|x \right\|^{2}}y$

Answer 1

The parameter $\beta_\text{opt}$ is the value that minimize $g(\beta)=\|y-\beta x\|^2=(y-\beta x)^T(y-\beta x)=\beta^2x^Tx-2\beta x^Ty+y^Ty$. And of course, the parabola $ax^2+bx+c$ achieves its minimum at $x=-\frac{b}{2a}$ for $a>0$. That is: $$\beta_\text{opt}=-\frac{-2x^Ty}{2x^Tx}=\frac{x^T}{\|x\|^2}y$$