Linear Maps and Geometry

Preserve combinations

A map $T$ is linear when $T(a\mathbf u+b\mathbf v)=aT(\mathbf u)+bT(\mathbf v)$ for all vectors and scalars. In finite-dimensional coordinate spaces every linear map can be represented by a matrix. The columns are the images of the standard basis vectors.

Linearity requires $T(0)=0$ , but that condition alone is insufficient: $T(x)=x^2$ maps zero to zero and is not linear. Translations by a nonzero vector are affine maps, not linear maps.

Worked example: build a matrix

Suppose $T(1,0)=(2,1)$ and $T(0,1)=(-1,3)$ . Then

A=\begin{pmatrix}2&-1\\1&3\end{pmatrix},\quad T(x,y)=(2x-y,x+3y).

For input $(3,2)$ , the output is $(4,9)$ . This follows by writing $(3,2)=3(1,0)+2(0,1)$ and applying preservation of combinations.

Explore

A matrix transforms the unit square

Try this. Set scale to 1 and vary shear: area stays 1 while the shape changes. Then set scale to 0 and −1 to compare collapse with orientation reversal.

Horizontal scale1.5Horizontal shear0.5

A(x,y) = (1.5x + 0.5y, y). Dashed: original unit square. Green: A(1,0) = (1.5, 0). Orange: A(0,1) = (0.5, 1). Determinant = 1.5; area = 1.5. Orientation preserved.

The explorer scales the horizontal direction and shears by an amount proportional to height. Watch both basis arrows and the transformed unit square. At zero horizontal scale, the square collapses to a line and the map loses invertibility.

Worked example: kernel and image

The projection $T(x,y)=(x,0)$ has matrix $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ . Its kernel is the vertical axis: all those inputs disappear. Its image is the horizontal axis: those are the reachable outputs. A map is one-to-one exactly when its kernel is zero, and onto its stated codomain exactly when its image is the whole codomain.

For finite-dimensional maps between spaces of equal dimension, one-to-one and onto are equivalent. With unequal dimensions they are different conditions, so always identify domain and codomain before using these words.

Practice

Is $T(x,y)=(x+1,y)$ linear?
Write the matrix that rotates the plane counterclockwise by $90^\circ$ .
Find the kernel of $T(x,y,z)=(x,y)$ .

Show worked solutions

No: $T(0,0)=(1,0)$ .
Basis images are $(0,1)$ and $(-1,0)$ , so the matrix is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .
All $(0,0,z)$ ; a basis is $(0,0,1)$ .