 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-i/1/c/iii/solution

The necessary and sufficient condition is

ker T = {0} .

ST = id_{V}

and

v \in ker T

, then

v = S (T v) = S (0) = 0

For sufficiency, let

e_{1}, e_{2}

be a basis of

V

and suppose the kernel of a linear map is trivial. The vectors

T e_{1}, T e_{2}

a T e_{1} + b T e_{2} = 0

implies

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

. Extend them to a basis

T e_{1}, T e_{2}, w_{3}

W

S

S (T e_{1}) = e_{1}, S (T e_{2}) = e_{2}, S (w_{3}) = 0.

Then

ST

agrees with the identity operator on a basis, hence on all of

V

. Thus a left inverse exists precisely when

T

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