11.4 Systems and transformations

In Module 2.2, we solved several linear equations at once. In Module 11.3, each matrix row produced one output from the same inputs. Put those ideas together: a system of equations can be written as one matrix calculation. The same calculation also tells us how a matrix moves every input vector to an output vector. These are two ways to read $Ax=b$.

This is intermediate linear algebra. OpenStax's lesson on matrix systems develops solving methods. Review matrix–vector multiplication (Module 11.3) first.

A basic idea: equations as rows, transformations as columns

The system $2x+y=5$, $x-y=1$ becomes

$$\underbrace{\begin{bmatrix}2&1\\1&-1\end{bmatrix}}_{A} \underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\text{unknown}} =\underbrace{\begin{bmatrix}5\\1\end{bmatrix}}_{b}.$$

Each row encodes one equation, so writing $Ax=b$ packages the system from Module 2.2; it does not change its solutions. For the geometric reading, ask what $A$ does to a point $(x,y)$: $A(x,y)^{\mathsf T}$ is its new position. The first column of $A$ is where $(1,0)$ lands and the second is where $(0,1)$ lands. These two unit arrows are called the standard basis vectors. Knowing where they land tells us where every point lands, because $(x,y)=x(1,0)+y(0,1)$.

A quick tour of properties

For $A$ of shape $m\times n$, $Ax=b$ has $m$ equations and $n$ unknowns. A system can have exactly one solution, no solution, or infinitely many. A linear transformation $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(cu)=cf(u)$; therefore it always sends the zero vector to zero. An added offset would make the map affine, not linear.

Application 1: solve two equations

Solve $2x+y=5$ and $x-y=1$.

1. From the second equation, $y=x-1$.
2. Substitute into the first: $2x+(x-1)=5$, so $3x=6$ and $x=2$.
3. Then $y=1$. Check: $A(2,1)^{\mathsf T}=(5,1)^{\mathsf T}$.

Application 2: shear a point

Let $S=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Its columns show $S(1,0)=(1,0)$ and $S(0,1)=(1,1)$. For $(2,1)$,

$$S\begin{bmatrix}2\\1\end{bmatrix} =2\begin{bmatrix}1\\0\end{bmatrix} +1\begin{bmatrix}1\\1\end{bmatrix} =\begin{bmatrix}3\\1\end{bmatrix}.$$

The basis-vector view and the row calculation agree.