Direct Sum

4. Direct summand

Let V be a vector space. Let H be a subspace of V. A direct summand of H in V is a subspace K such that V = HK.

Example From this example, the subspace of skew-symmetric matrices is a direct summand of the subspace of symmetric matrices.

From this example, the subspace of odd functions is a direct summand of the subspace of even functions.

Example As explained in this example, the vectors (3, 1, 2), (1, -1, 2), (-1, 1, 1) form a basis of R3. Let

H = span{(3, 1, 2), (1, -1, 2)}, K = span{(-1, 1, 1)}.

Then the fact that the three vectors span R3 implies H + K = R3. Moreover, an argument similar to this example shows that the sum H + K is a direct one. Therefore K is a direct summand of H (and H is also a direct summand of K).

The argument in the example above can be extended. In general, if a basis {b1, b2, ..., bn} of V is divided into two disjoint parts {b1, b2, ..., bk} and {bk+1, bk+2, ..., bn}, then span{bk+1, bk+2, ..., bn} is a direct summand of span{b1, b2, ..., bk}. The idea can be used to prove the existence of the direct summand.

Any subspace of a finite dimensional vector space has a direct summand.

Proof Let H be a subspace of a finite dimensional vector space V. Let B = {b1, b2, ..., bk} be a basis of H. If we can extend B to a basis B' = {b1, b2, ..., bn} be a basis of V, then by the remark made before, the subspace K = span{bk+1, bk+2, ..., bn} is a direct summand of H = span{b1, b2, ..., bk}.

If V = H = span{b1, b2, ..., bk}, then B is already a basis of V. Otherwise, we can find bk+1H. By this result, b1, b2, ..., bk, bk+1 are linearly independent.

If V = H1 = span{b1, b2, ..., bk, bk+1}, then we have extended B to a basis of V by adding one extra vector. Otherwise, we find another bk+2H1. Again, b1, b2, ..., bk, bk+1, bk+2 are linearly independent.

Again we ask whether V = H2 = span{b1, b2, ..., bk, bk+1, bk+2}. The process may continue, and we get larger and larger collection of linearly independent vectors. Since V is finite dimensional, the process will eventually stop at V = Hn-k = span{b1, b2, ..., bk, bk+1, bk+2, ..., bn}. Then B' = {b1, b2, ..., bk, bk+1, bk+2, ..., bn} is a basis obtained by extending B.

The idea of the proof is by extending B one vector at a time (and hence extending H one dimension at a time) until V is "filled up". For another approach and some concrete examples of extending bases, see this exercise.