### Direct Sum

##### 2. Direct sum of subspaces

Let H1, H2, ..., Hk be subspaces of V. Then

S(v1, v2, ..., vk) = v1 + v2 + ... + vk: H1H2⊕...⊕HkV

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 = H1H2⊕...⊕Hk.

Thus a sum of subspaces is a direct sum simply means that H1 + H2 + ... + Hk is isomorphic to H1H2⊕...⊕Hk in a natural way. By the definition, the condition for a sum to be direct is

v1 + v2 + ... + vk = 0, viHiv1 = v2 = ... = vk = 0.

In the special case Hi = span{ui} are 1-dimensional lines (ui0), 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): vH1H2}. Therefore

H1 + H2 is a direct sum ⇔ H1H2 = {0}. Example Consider subspaces

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 = 0c1 = 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 HK consists of solutions of the system Ax = 0. From this example, we know the solution has to be trivial. Therefore HK = {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 = {fF(R): f(t) = f(-t)} of F(R). By the similar reason, the odd functions also form a subspace K = {fF(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 HK = F(R).