1 If dim(V) >dim(W), then T is not injective. Linear algebra An injective linear map between two finite dimensional vector spaces of the same dimension is surjective. Let $$T : V \rightarrow W$$ be a linear map between vector spaces. Recall that the composition TS is defined by (TS)(x) = T(S(x)). Let U, V, and W be vector spaces over F where F is R or C. Let S: U -> V and T: V -> W be two linear maps. Example 5. By the rank-nullity theorem, the dimension of the kernel plus the dimension of the image is the common dimension of V and W, say n. By the last result, T is injective 2 If dim(V)