pick a basis B = {b_{1},...,b_{n}} for V.

thus we can write any v in V as v = c_{1}b_{1}+...+c_{n}b_{n}.

define T(c_{1}b_{1}+...+c_{n}b_{n}) = c_{n}b_{1}.

T is non-zero because T(b_{n}) = b_{1}≠ 0, since b_{1}is a basis vector.

but T^{2}(v) = T(T(v)) = T(T(c_{1}b_{1}+...+c_{n}b_{n})) = T(c_{n}b_{1}) = 0

(because c_{n}b_{1}= c_{n}b_{1}+ 0b_{2}+....+0b_{n}).

basically we are using the linear transformation that has the matrix (relative to the basis B):