Proof that the composition of linear maps is linear

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This proof is mostly bookkeeping, which makes it a good Proof Builder page. The embed keeps the two required checks separate: addition first, scalar multiplication second.

What has to be shown

Let $S : U \to V$ and $T : V \to W$ be linear. The composition $T \circ S$ sends a vector $u \in U$ first through $S$ , then through $T$ .

To prove $T \circ S$ is linear, you must check the two defining properties: it preserves vector addition and scalar multiplication.

The addition check

Start with $(T \circ S) (u + v)$ . By definition this is $T (S (u + v))$ .

Linearity of $S$ rewrites the inside as $S (u) + S (v)$ . Then linearity of $T$ separates the outside into $T (S (u)) + T (S (v))$ , which is exactly $(T \circ S) (u) + (T \circ S) (v)$ .

The scalar check

The scalar step follows the same pattern. First $S (c u) = c S (u)$ because $S$ is linear.

Then $T (c S (u)) = c T (S (u))$ because $T$ is linear. That leaves $(T \circ S) (c u) = c (T \circ S) (u)$ .

Common Pitfall

The proof does not say every composition of functions is linear. Both maps must already be linear, and the output space of the first map must match the input space of the second.

Try a Variation

Try composing $S (x, y) = (x + y, y)$ with $T (a, b) = (2 a - b, a)$ . Compute $(T \circ S) (x, y)$ , then check the two linearity conditions directly.

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