Canonical Forms Linear Algebra Notes

2.13 (Definition) Let V be a vector space over F and V1, V2 . . . , Vk
be subspaces of V. We say that V is direct sum of V1, V2, . . . , Vk,
if each element x ∈ V can be written uniquely as
with ωi ∈ Vi.
Equivalently, if
x = ω1 + ω2 + · · · + ωk
1. V = V1 + V2 + · · · + Vk, and
2. ω1 + ω2 + · · · + ωk = 0 with ωi ∈ Vi implies that ωi = 0 for
i = 1, . . . , k.
If V is direct sum of V1, V2 . . . , Vk then we write
V = V1 ⊕ V2 ⊕ · · · ⊕ Vk.
Following is a proposition on direct sum decomposition.
2.14 (Proposition) Let V be a vector space over F with dim(V ) =
n < ∞. Let V1, V2 . . . , Vk be subspaces of V Then
V = V1 ⊕ V2 ⊕ · · · ⊕ Vk
if and only if V = V1 + V2 + · · · + Vk and
dim(V ) = dim(V1) + dim(V2) + · · · + dim(Vk).
Proof. (⇒): Obvious.
(⇐): Let Ei = {eij : j = 1, . . . , di} be basis of Vi. Let E = {eij :
j = 1, . . . , di; i = 1, . . . , k}. Since V = V1 + V2 + · · · + Vk, we have
V = SpanE. Since dim(V ) = cardinlity(E), we have E forms a
basis of V. Now it follows that if ω1 + · · · + ωk = 0 with ωi ∈ Wi then
ωi = 0 ∀i. This completes the proof.
Now we restate the final theorem 2.12 in terms of direct sum.
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