 Solution

ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2021/1/a/i/solution

V / U

identifies two vectors

v, w

precisely when

v - w \in U

. Since

U

T (U) \subseteq U

. Thus

v + U = w + U ⟹ v - w \in U ⟹ T (v) - T (w) = T (v - w) \in U .

Consequently

T (v) + U = T (w) + U

: the output does not depend on the chosen representative. This proves that the induced linear map on a quotient vector space

\overline{T} (v + U) = T (v) + U

is well-defined. It is a linear map because, for scalars

a, b

\overline{T} (a (v + U) + b (w + U)) = T (a v + b w) + U = a \overline{T} (v + U) + b \overline{T} (w + U) .

 New to topics? Read the docs here!