A map takes a vector and returns a vector. It is called linear when it respects the two operations: adding first and mapping gives the same as mapping first and adding, and the same for scaling.

Converting a price list to another currency is linear. Two orders converted separately and then added cost the same as the orders added and then converted.

Before the map: two vectors and their sum, walked tip to tail.

After a quarter turn. The whole triangle has turned with the plane, so the arrow that was the sum is still the sum: mapping then adding and adding then mapping arrive at one place. A map for which this fails, however sensible it looks, is not linear.

A map $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ is linear when it satisfies both conditions below for every pair of vectors and every scalar. Both are needed: a map may respect one and break the other. Squaring each component respects neither, and adding a fixed non-zero vector respects neither.

• $T\mathbf{0}=\mathbf{0}$: take $c=0$ in the second condition.
• A map with $T\mathbf{0}\neq\mathbf{0}$ is therefore not linear, whatever else it does.

Unfold both sides using the definition of the product as a weighted sum of columns. Adding two vectors adds their weights position by position, and the rule for scaling by a sum turns the pair of terms on each column into one. Scaling a vector multiplies every weight, which the rule for scaling a combination hands straight back out.

Unfold the product: the weight on column j is component j of $\mathbf{u}+\mathbf{v}$, which is $u_j+v_j$.

Each term splits by the rule for scaling by a sum, from the lesson on adding and scaling vectors.

Gather the terms carrying $\mathbf{u}$ and those carrying $\mathbf{v}$; each group is a product by definition.

Scaling the vector multiplies every weight by c, and c comes back out of the whole combination.

Both required conditions hold for every $\mathbf{u},\mathbf{v}$ and every c, which is the definition.

$T(\mathbf{u}+\mathbf{v})=T\mathbf{u}+T\mathbf{v}$ and $T(c\mathbf{u})=cT\mathbf{u}$

Neither line implies the other, so neither may be skipped.

$T\mathbf{x}=\mathbf{x}+\mathbf{p}$ gives $T\mathbf{0}=\mathbf{p}$

With $\mathbf{p}\neq\mathbf{0}$ the zero vector has moved, so $T$ is not linear.

$T(x,y)=(x^2,y)$ gives $T(2,0)=(4,0)$

Doubling the input should double the output, and it quadrupled the first component.

Split the sum with the first condition, then pull out each scalar with the second.