What Linear Algebra Is Really About

A vector has two faces. It is an arrow with a direction and a length, and it is a list of numbers, its coordinates in some basis. Linear algebra is the study of the maps that respect this structure, and its central surprise is that one such map and one grid of numbers, a matrix, are the same object seen two ways.

Linear maps

Fix vector spaces and call a function $T$ linear when it commutes with the two operations vectors have, addition and scaling:

$$T(a\,x + b\,y) = a\,T(x) + b\,T(y), \qquad a, b \in \mathbb{R}.$$

That single rule is strong. Write any vector in the standard basis $e_1 = (1,0)$, $e_2 = (0,1)$ as $x = x_1 e_1 + x_2 e_2$. Applying $T$ and using linearity once for each term,

$$T(x) = T(x_1 e_1 + x_2 e_2) = x_1\,T(e_1) + x_2\,T(e_2).$$

So $T$ is pinned down everywhere by the two output vectors $T(e_1)$ and $T(e_2)$. Know where the basis lands and you know the whole map.

A matrix is a linear map

Stack those two images as the columns of a matrix,

$$A = \begin{bmatrix} T(e_1) & T(e_2) \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix},$$

and the formula above is exactly the matrix-vector product:

$$A x = x_1 \begin{bmatrix} a \\ c \end{bmatrix} + x_2 \begin{bmatrix} b \\ d \end{bmatrix} = \begin{bmatrix} a x_1 + b x_2 \\ c x_1 + d x_2 \end{bmatrix}.$$

Reading it backward gives the fact worth memorizing: $A e_1$ is the first column and $A e_2$ is the second column, because $A e_1 = 1\cdot(\text{col }1) + 0\cdot(\text{col }2)$. A matrix is a table of where the basis vectors go. Multiplying by $A$ is applying the map $T$.¹

A matrix bends the grid to its columnsinteractive

a = 2.0b = 1.0c = 0.0d = 3.0

A e₁ (col 1)

(2.0, 0.0)

A e₂ (col 2)

(1.0, 3.0)

A x

(3.0, 3.0)

det = ad − bc

6.00 − 0.00 = 6.00

Area scale = |det| = 6.00. The unit square becomes a parallelogram of that area.

A worked example

Take

$$A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}.$$

The columns tell us immediately where the basis lands,

$$A e_1 = \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \qquad A e_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix},$$

and any other vector follows by linear combination. For $x = (1, 1)$,

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

We never needed a new rule, only $A e_1$, $A e_2$, and linearity. The same computation done componentwise reads $2(1) + 1(1) = 3$ and $0(1) + 3(1) = 3$.

The determinant is area scaling

The unit square with sides $e_1$ and $e_2$ has area $1$. Under $A$ it becomes the parallelogram with sides $A e_1$ and $A e_2$, and the (signed) area of that parallelogram is the determinant,

$$\det A = ad - bc.$$

For the example, $\det A = 2\cdot 3 - 1\cdot 0 = 6$: every region has its area multiplied by $6$. The sign carries orientation. If $\det A < 0$ the map flips the plane over (the demo's parallelogram turns inside out), and if $\det A = 0$ the two columns are parallel, the square collapses onto a line, and $A$ crushes the whole plane into a lower dimension. That last case is precisely when $A$ has no inverse, which is the thread Part 1 picks up.

Footnotes

1. Strang, Introduction to Linear Algebra (5th ed., 2016). ↩