Scalar multiplication: stretch, squeeze, and flip

Scaling a vector by $c$ rescales its length by $|c|$ and flips direction when $c\lt 0$. The chapter stresses visualization: number lines along the span of the vector interpret scalar multiplication as motion on a one-dimensional shadow .

When $c\gt 1$, the arrow lengthens by factor $c$ while staying on the same line through the origin. When $0\lt c\lt 1$, it shortens but keeps direction. Multiplying by $-1$ sends $\mathbf{v}$ to a vector pointing the opposite direction with the same magnitude.

This is the first appearance of linearity in one slot: fix the vector, scaling is homogeneous. Scalar multiplication distributes across vector addition because each coordinate is a real number obeying distributivity: $c(\mathbf{u}+\mathbf{v})=c\mathbf{u}+c\mathbf{v}$ componentwise .

Edge case: scaling the zero vector by any $c$ still yields $\mathbf{0}$. The zero vector has no preferred direction, yet it participates in every linear combination.

Check your understanding. The tasks below rest on these ideas: Correct: a positive scalar greater than $1$ stretches the arrow along its line by that factor while keeping its direction. Not quite: it cannot shrink or keep the same length for $c\gt 1$, and only a negative scalar would reverse direction. Correct: the factor $|-1|=1$ keeps the length while the negative sign flips the direction. Not quite: negation does not rotate to perpendicular, does not collapse a nonzero vector to $\mathbf{0}$, and leaves the length unchanged rather than halving it. Correct: scaling and adding both act coordinate by coordinate, so real-number distributivity gives the identity in every coordinate. Not quite: distributivity needs no angle preservation, no orthogonal basis, and is unrelated to how lengths combine. Correct: the negative sign flips the direction and the magnitude scales by $|c|$. Not quite: a negative scalar must reverse, not preserve, direction; scaling is not rotation; and it never produces a perpendicular vector.