 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/1/a/solution

im T = {T (v) : v \in V} \subseteq W, ker T = {v \in V : T (v) = 0} \subseteq V .

A linear map sends zero to zero, since

T (0) = T (0 + 0) = T (0) + T (0)

. Thus both displayed sets contain zero. If

w_{1} = T (v_{1})

and

w_{2} = T (v_{2})

belong to the image of a linear map, then for any real scalars

a, b

a w_{1} + b w_{2} = T (a v_{1} + b v_{2}) \in im T .

v_{1}, v_{2}

T (a v_{1} + b v_{2}) = a T (v_{1}) + b T (v_{2}) = 0,

a v_{1} + b v_{2} \in ker T

. Containing zero and being closed under linear combinations is the vector subspace criterion. Hence the image of a linear map is a vector subspace of

W

V

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