How is the "two intersecting lines" picture of a system of two linear equations related to the "sum of two vectors" picture?

Question

Suppose I have the set of linear equations

$$\begin{array}{cc}ax+by &= e \\ cx+dy &= f\end{array}$$

One picture I can draw to represent this system consists of two lines. The first is a line through $(0,e/b)$ and $(e/a,0)$, representing all solutions to the first equation, and the second is a line through $(0,f/d)$ and $(f/c,0)$ representing all solutions to the second equation. The simultaneous solution is where the lines intersect.

A second picture is that I draw the vector $(a,c)$ and imagine myself free to scale its length by a factor $x$. Then I draw the vector $(b,d)$ and scale its length by a factor $y$. The simultaneous solution to the equations is the set $(x,y)$ such that when I scale the vectors by $x$ and $y$, they add to the vector $(e,f)$.

It is clear to me that both of these are valid representations of the system of equations, but their relationship to each other seems a little mystical. If you described each picture to me independently, I wouldn't see them as being different representations of the same thing unless I turned them both into the same algebraic system of equations. Is there a more direct geometric way to see that these two representations are describing the same problem?

I built a quick interactive picture of these representations on geogebra here if that is helpful.

Answer 1

I would say that your two points of view correspond to whether you want to investigate a linear map from the "row" point of view or from the "column" point of view. Both are important and the interplay between them is often what makes linear algebra very interesting. Let me elaborate. Define a linear map $T \colon \mathbb{R}^2_{\operatorname{col}} \rightarrow \mathbb{R}^2_{\operatorname{col}}$ by

$$ T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}. $$

We want to find $(x,y)^T$ such that

$$ T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix}. $$

Then we can proceed in one of two ways:

1. Let us focus on the rows of $A$. Each row of $A$ corresponds to a coordinate of $T$ as $$ T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}. $$ In fact, we can think of each row as defining a linear functional $\varphi_i$ on $\mathbb{R}^2_{\operatorname{col}}$ as follows: $$ \varphi_1 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = ax + by, \\ \varphi_2 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = cx + dy $$ and then $$ T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \varphi_1 \begin{pmatrix} x \\ y \end{pmatrix} \\ \varphi_2 \begin{pmatrix} x \\ y \end{pmatrix} \end{pmatrix}. $$ In order to solve the system, we need to solve $$\varphi_1 \begin{pmatrix} x \\ y \end{pmatrix} = e, \varphi_2 \begin{pmatrix} x \\ y \end{pmatrix} = f$$ and then intersect the resulting solution sets. This is the "system of equations" point of view.
2. Let us focus on the columns of $A$. The map $T$ is linear and so we can write $$ T \begin{pmatrix} x \\ y \end{pmatrix} = T \left( x \begin{pmatrix} 1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right) = x T \begin{pmatrix} 1 \\ 0 \end{pmatrix} + y T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = x \begin{pmatrix} a \\ c \end{pmatrix} + y \begin{pmatrix} b \\ d \end{pmatrix}. $$ This shows us that the image of $T$ consists of linear combinations of the columns of $A$. From this point of view, to solve $T$ we need to find $x,y$ such that the vector $$ \begin{pmatrix} e \\ f \end{pmatrix} $$ is a linear combination of the columns of $A$ with coefficients $x,y$.

Answer 2

At the risk of adding confusion to the subject:

Adding a third variable $z$ the problem becomes projective (the affine subspace $z=1$ corrsponds to the original problem) and we may define a linear map $T: {\Bbb R}^3 \rightarrow {\Bbb R}^2$ by setting: $$ T \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} a & b & -e \\ c & d & -f \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) $$ Which may then be written in the two different ways:

$$ \left( \begin{matrix} a \\ c \end{matrix} \right) x + \left( \begin{matrix} b \\ d \end{matrix} \right)y + \left( \begin{matrix} -e \\ -f \end{matrix} \right)z \ \ \ \ \mbox{or} \ \ \ \ \left( \begin{matrix} ax+by-ez \\ cx +dy -fz \end{matrix} \right)$$ We are trying to describe the kernel of $T$. For $(x,y,z)$ in the kernel, the first expression describes the linear dependency between three two-dimensional vectors, whereas the last the simultaneous vanishing of two linear forms on ${\Bbb R}^3$, which geometrically is the intersection of two planes. If a solution admits $z=1$ then you have a solution of your original problem. The way of writing illustrates that geometrically we are dealing with two distinct problems, one is in 2 dimensions, the other in 3. It is somehow 'normal' that it is difficult to give an intuitive geometric link between the two. Note that in your descriptions, although both are in two dimensions, one is affine, the other vectorial.