not linearly independent, and let for some . Thus, owing to the linearity of , it follows that This is a contradiction to being a basis, unless all are equal to zero. This shows that is linearly independent, and more specifically that it is a basis for . To summarize, we have , a basis for , and , a basis for . Finally we may state that This concludes our proof. Second proof Let with linearly independent columns (i.e. ). We will show that: To do this, we will produce a matrix whose columns form a basis of the null space of . Without loss of generality, assume that the first columns of are linearly independent. So, we can write where with linearly independent column vectors, and , each of whose columns are linear combinations of the columns of . This means that for some (see rank factorization) and, hence, Let where is the identity matrix. We note that satisfies Therefore, each of the columns of are particular solutions of . Furthermore, the columns of are linearly independent because will imply for : Therefore, the column vectors of constitute a set of linearly independent solutions for . We next prove that any solution of must be a linear combination of the columns of . For this, let be any vector such that . Note that since the columns of are linearly independent, implies . Therefore, This proves that any vector that is a solution of must be a linear combination of the special solutions given by the columns of . And we have already seen that the columns of are linearly independent. Hence, the columns of constitute a basis for the null space of . Therefore, the nullity of is . Since equals rank of , it follows that . This concludes our proof. Reformulations and generalizations This theorem is a statement of the first isomorphism theorem of algebra for the case of vector spaces; it generalizes to the splitting lemma. In more modern language, the theorem can also be phrased as saying that each short exact sequence of vector spaces splits. Explicitly, given that is a short exact sequence of vector spaces, then , hence Here R plays the role of im T and U is ker T, i.e. In the finite-dimensional case, this formulation is susceptible to a generalization: if is an exact sequence of finite-dimensional vector