math111_logo Direct Sum

3. Properties of direct sum

Recall that the sum of spans of single vectors being a direct sum is equivalent to the linear independence. The following is an extension of the observation, which complements this result about spanning vectors in a sum.

Let H1 + H2 + ... + Hk be a direct sum of subspaces of V. Let Bi = {vi1, vi2, ...} be linearly independent vectors in Hi. Then B1B2∪...∪Bk = {vij: all i, j} are linearly independent.

Proof For the case k = 3 and B1 = {v11, v12}, B2 = {v21, v22}, B3 = {v31, v32, v33} are linearly independent vectors in H1, H2, H3, we will prove that B = {v11, v12, v21, v22, v31, v32, v33} is linearly independent. The general case is similar.

Let a linear combination of vectors of B be trivial:

(c11v11 + c12v12) + (c21v21 + c22v22) + (c31v31 + c32v32 + c33v33) = v1 + v2 + v3 = 0.

We need to conclude that all the seven coefficients are zero. Since

v1 = c11v11 + c12v12H1, v2 = c21v21 + c22v22H2, v3 = c31v31 + c32v32 + c33v33H3,

the fact that H1 + H2 + H3 is a direct sum implies that v1 = v2 = v3 = 0. Then by c11v11 + c12v12 = v1 = 0 and B1 = {v11, v12} being linearly independent, we have c11 = c12 = 0. Similarly, we have c21 = c22 = 0 and c31 = c32 = c33 = 0.

In the special case Bi is a basis of Hi, we conclude the ⇒ direction of the following.

H1 + H2 + ... + Hk is a direct sum ⇔ dim(H1 + H2 + ... + Hk) = dimH1 + dimH2 + ... + dimHk.

We remark that by this exercise, we have dim(H1 + H2 + ... + Hk) ≤ dimH1 + dimH2 + ... + dimHk in general.

Proof For the ⇐ direction, we again take a basis Bi of Hi. Then B = B1B2∪...∪Bk contains dimH1 + dimH2 + ... + dimHk number of vectors and spans H1 + H2 + ... + Hk. In case the equality dim(H1 + H2 + ... + Hk) = dimH1 + dimH2 + ... + dimHk holds, the number of vectors in B is the same as the dimension of H1 + H2 + ... + Hk. By this result, B is a basis of H1 + H2 + ... + Hk.

Now for the special case k = 3 and B1 = {v11, v12}, B2 = {v21, v22}, B3 = {v31, v32, v33}, the linear independence of B means that

(c11v11 + c12v12) + (c21v21 + c22v22) + (c31v31 + c32v32 + c33v33) = 0

implies all seven coefficients are zero. Equivalently, for the arbitrary (becase B1, B2, B3 are bases) vectors

v1 = c11v11 + c12v12H1, v2 = c21v21 + c22v22H2, v3 = c31v31 + c32v32 + c33v33H3,

v1 + v2 + v3 = 0 implies v1 = v2 = v3 = 0. This means exactly that H1 + H2 + H3 is a direct sum.


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