What do we gain by representing solutions of Systems of Linear Equations as Parametric Equations?

Question

Consider the following system of linear equations: \begin{aligned} 4x − 2y &= 1 \\ 16x − 8y &= 4 \end{aligned} We can eliminate $x$ from the second equation by adding $(−4) \times$ the first equation to the second. This yields the simplified system \begin{aligned} 4x − 2y &= 1 \\ 0 &= 0 \\ \end{aligned} The second equation does not impose any restrictions on $x$ and $y$ and hence can be omitted. Thus, the solutions of the system are those values of $x$ and $y$ that satisfy the single equation $$ 4x − 2y = 1 $$ One way to describe the solution set is to solve this equation for $x$ in terms of $y$ to obtain $x = \frac{1}{4} + \frac{1}{2}y$ and then assign an arbitrary value $t$ (called a parameter) to $y$. This allows us to express the solution by the pair of equations (called parametric equations) $$ x = \frac{1}{4} + \frac{1}{2}t, \qquad y = t $$ My Question is: What do we gain by representing the 2nd-last equation in a parametric way? Are there any obvious advantages that I am not seeing ? I have some understanding of parametric equations but I am unable to relate the solution of linear system to the concept of parametric equations ? or stated otherwise how do they relate to each other ?

Answer 1

In your example the solution of the linear system is a line.

Here are some representations of a line in 2D I use: $$ y = a x + b \\ a x + b y + c = 0 \\ n \cdot (x, y)^\top = d \\ (x, y)^\top = (p_x, p_y)^\top + t\, (d_x, d_y)^\top \\ (1-\lambda) u_1 + \lambda u_2 $$ It depends on the problem, what form I use. E.g. the parametric case is interesting for physical problems, where we are interested in the movement of a point on the line.

If we move to three dimensions, I would need to update that list, e.g. using the intersection of linear equations / affine planes, I would need two of them, and they need to be linear independent: $$ n_1 \cdot (x, y, z)^\top = d_1 \\ n_2 \cdot (x, y, z)^\top = d_2 \\ $$ But I could keep the parametric forms more or less: $$ (x, y, z)^\top = (p_x, p_y, p_z)^\top + t \, (d_x, d_y, d_z)^\top \\ (1-\lambda) u_1 + \lambda u_2 $$

Answer 2

The following is more of a philosophical nature. It does not refer to your specific set of equations.

An equation $$\Phi(x)=0,\quad x\in X,\tag{1}$$ which might also be a system of equations, implicitly defines a solution set $$S:=\bigl\{x\in X\bigm|\Phi(x)=0\bigr\} \ \subset X\ .$$ For any trial point $x\in X$ it is easy to check whether it belongs to $S$ or not, but we don't have a birds eye view over the set $S$. Solving the equation means producing an explicit representation of $S$ in the form of a "list". This list may be empty (if there are no solutions), or consist of exactly one entry, which is then called the solution of $(1)$. As long as $S$ is a finite set we can present it in the form $S=\{a, b,\ldots, p\}$. If the set $S$ is infinite we have to resort to a "production scheme" consisting of an index set $I$ and a function $f:\>I\to X$ that outputs for each $\iota\in I$ an element $x_\iota\in S$ in an essentially bijective way. Such an $f$ is then called a parametric representation of $S$.

In your example we can write $S$ in the form $S=f({\mathbb R})$ with $$f:\quad{\mathbb R}\to{\mathbb R}^2,\qquad t\mapsto\left({1\over4}+{t\over2},\>t\right)\ .\tag{1}$$ Here $X={\mathbb R}^2$, the index set is the $t$-axis ${\mathbb R}$, and the $f$ in $(1)$is the "production scheme" for $S$.

Answer 3

Comparing

$$x(y) = \frac{1}{4} + \frac{1}{2}y$$

vs

$$x(t) = \frac{1}{4} + \frac{1}{2}t, \qquad y(t) = t$$

In the first you represent one of the original variables in terms of the other, describing your solution set with the mapping

$$y \to (x(y),y)$$

In the second it's not much different other than using an arbitrary variable, $ \ t$:

$$t \to (x(t),y(t))=\left(\frac{1}{2}t+\frac{1}{4},t\right)$$

Which only will yield the same set of $(x,y)$'s just like above. But here you can make the substitution

$$\quad t\to 2t-\frac{1}{2} \quad $$

for example, leading you to another description of your solution set:

$$t\to (x_2(t),y_2(t))=\left(t,2t-\frac{1}{2}\right)$$

Which again yields that very same set of $(x,y)$'s.

Using an arbitrary parameter is a degree of freedom that, say, the first solution set description doesn't have. It gives you a way of describing your solution set in more than one way, which will come into play later in your linear algebraic studies, and in higher dimensions.