Direct Sum Direct SumLet H1, H2, ..., Hk be subspaces of V. Then
S(v1, v2, ..., vk) = v1 + v2 + ... + vk: H1⊕H2⊕...⊕Hk → V
is a linear transformation with the subspace H1 + H2 + ... + Hk as the range.
The sum H1 + H2 + ... + Hk of subspaces of a vector space V is a direct sum if the natural linear transformation S is one-to-one. If we also have V = H1 + H2 + ... + Hk, then we say V is a direct sum of the subspaces and denote V = H1⊕H2⊕...⊕Hk.
Thus a sum of subspaces is a direct sum simply means that H1 + H2 + ... + Hk is isomorphic to H1⊕H2⊕...⊕Hk in a natural way. By the definition, the condition for a sum to be direct is
v1 + v2 + ... + vk = 0, vi ∈ Hi ⇒ v1 = v2 = ... = vk = 0.
In the special case Hi = span{ui} are 1-dimensional lines (ui ≠ 0), we have vi = ciui, and the condition above becomes the linear independence of v1, v2, ..., vk. Thus direct sum generalizes the concept of linear independence. Moreover, in case k = 2, we have kernelS = {(v, -v): v ∈ H1∩H2}. Therefore
H1 + H2 is a direct sum ⇔ H1∩H2 = {0}.
H1 = span{(1, -1, 0, 1)}, H2 = span{(3, -1, 4, 3), (0, 1, 4, -2)}, H3 = span{(1, 1, 3, 0)}
of R4. For any vectors
v1 = c1(1, -1, 0, 1), v2 = c2(3, -1, 4, 3) + c3(0, 1, 4, -2), v3 = c4(1, 1, 3, 0)
in the three subspaces, we have
v1 + v2 + v3 = c1(1, -1, 0, 1) + c2(3, -1, 4, 3) + c3(0, 1, 4, -2) + c4(1, 1, 3, 0).
By the row operations in this example (look at columns 1, 2, 4, 6), the four vectors are linearly independent. Therefore
v1 + v2 + v3 = 0 ⇒ c1 = c2 = c3 = c4 = 0 ⇒ v1 = v2 = v3 = 0,
and we conclude H1 + H2 + H3 is a direct sum.
In general, if a basis of V is divided into a disjoint union of subsets, then the subspaces spanned by the subsets form a direct sum of V.
Example Consider the matrix
| A = [ | 1 | 3 | 0 | ]. |
| -1 | -1 | 1 | ||
| 0 | 4 | 4 | ||
| 1 | 3 | -2 |
Using first two rows as the coefficient matrix, we have a kernel subspace
H = {(x1, x2, x3): x1 + 3x2 = 0, -x1 - x2 + x3 = 0}.
Using the last two rows, we also have
K = {(x1, x2, x3): 4x2 + 4x3 = 0, x1 + 3x2 - 2x3 = 0}.
The intersection H∩K consists of solutions of the system Ax = 0. From this example, we know the solution has to be trivial. Therefore H∩K = {0}, and H + K is a direct sum.
Example Since the only symmetric and skew symmetric matrix is the zero matrix, the sum of the subspace of symmetric matrices and the subspace of skew-symmetric matrices is a direct one. In fact, by this exercise, we may say that the space of all n by n matrices is the direct sum of symmetric and skew symmetric matrices.
Example By this example (also see this exercise), the even functions form a subspace H = {f ∈ F(R): f(t) = f(-t)} of F(R). By the similar reason, the odd functions also form a subspace K = {f ∈ F(R): f(t) = -f(-t)}. Now for any function f(t) ∈ F(R), we construct
| fE(t) = | f(t) + f(-t) | , fO(t) = | f(t) - f(-t) |
. |
| 2 | 2 |
Then it is easy to verify that fE is an even function, fO is an odd function, and f = fE + fO. This means H + K = F(R). Moreover, if f is both even and odd, then we will have f(t) = f(-t) = -f(t), which implies f(t) = 0. Thus we conclude H⊕K = F(R).