### 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** = **H**⊕**K**.

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 **R**^{3}. Let

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

Then the fact that the three vectors span **R**^{3} implies **H** + **K** = **R**^{3}. 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 {`b`_{1}, `b`_{2}, ..., **b**_{n}} of `V` is divided into two disjoint parts {`b`_{1}, `b`_{2}, ..., **b**_{k}} and {`b`_{k+1}, `b`_{k+2}, ..., **b**_{n}}, then `span`{`b`_{k+1}, `b`_{k+2}, ..., **b**_{n}} is a direct summand of `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}}. 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` = {`b`_{1}, `b`_{2}, ..., **b**_{k}} be a basis of `H`. If we can extend `B` to a basis `B'` = {`b`_{1}, `b`_{2}, ..., **b**_{n}} be a basis of `V`, then by the remark made before, the subspace `K` = `span`{`b`_{k+1}, `b`_{k+2}, ..., **b**_{n}} is a direct summand of `H` = `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}}.

If `V` = `H` = `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}}, then `B` is already a basis of `V`. Otherwise, we can find `b`_{k+1} ∉`H`. By this result, `b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1} are linearly independent.

If `V` = `H`_{1} = `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1}}, then we have extended `B` to a basis of `V` by adding one extra vector. Otherwise, we find another `b`_{k+2} ∉`H`_{1}. Again, `b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1}, `b`_{k+2} are linearly independent.

Again we ask whether `V` = `H`_{2} = `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1}, `b`_{k+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` = `H`_{n-k} = `span`{`b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1}, `b`_{k+2}, ..., **b**_{n}}. Then `B'` = {`b`_{1}, `b`_{2}, ..., **b**_{k}, `b`_{k+1}, `b`_{k+2}, ..., **b**_{n}} 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.