Why do we parameterize our solutions/equations?

Question

I'm currently taking a linear algebra course, and we always have to parameterize our solution after finding it using rref. Why do we do this? Why can't we just leave the solution in terms of $x$, $y$, $z$ instead of converting it to $r$, $s$, $t$? What are the benefits of this? I've seen this in calculus too...

Answer 1

Let $T:\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation, and let $\mathbf{b}\in\mathbb{R}^m$. Parametrizing the solutions of $T(\mathbf{x})=\mathbf{b}$ is equivalent to finding a basis for the vector space $$\ker(T)=\{\mathbf{x}\in\mathbb{R}^n:T(\mathbf{x})=\mathbf{0}\}\subseteq\mathbb{R}^n$$ and finding a vector $\mathbf{x}_0\in\mathbb{R}^n$ with $T(\mathbf{x}_0)=\mathbf{b}$, so that $\{\mathbf{x}\in\mathbb{R}^n:T(\mathbf{x})=\mathbf{b}\}=\ker(T)+\mathbf{x}_0$.

Finding a basis for a vector space is both conceptually important (it tells you the dimension of the space and gives "coordinates" in which to visualize it) and computationally useful (it explicitly describes the vector space as consisting of all linear combinations of the basis elements, which eliminates the need for any further messing with equations). In particular, when it comes to finding a basis for a vector space of the form $\ker(T)$, your intuition should be that you are finding fundamentally "independent" solutions.

Another famous case of parametrizing solutions are Pythagorean triples. One could just stare at the set $C=\{(r,s)\in\mathbb{Q}^2:r^2+s^2=1\}$ and wonder what it's elements are, but if you've parametrized the solutions, you can confidently state that you can choose an arbitrary $t\in\mathbb{Q}$ and get a unique solution $(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2})\in C$, with the only solution missed thereby being $(-1,0)$. This again is massively important, conceptually and computationally.