Let A=[T]BVBW be the matrix representation of T:V→W over a
scalar field F, with dim(V)=n and dim(W)=m. Then
ker(A)={[v]BV:v∈ker(T)}
and
im(A)={[w]BW:w∈im(T)}.
Since a linear map is injective exactly when its kernel is trivial and
surjective exactly when its image is the codomain, these identities give the
coordinate tests ker(A)={0} and im(A)=Fm.
Here ker(A) and im(A) are sets of coordinate columns;
the displayed equalities identify them with the corresponding abstract
subspaces through the coordinate maps.
Consequently,
nullity(A)=nullity(T),rank(A)=rank(T).
Hence T is injective if and only if nullity(A)=0 and T is surjective if and only if rank(A)=m. If m=n, then T is bijective if and only if A is invertible.