Vector Space

A vector space is a set $V$ along with an addition on $V$ and a scalar multiplication on $V$ such that satisfies commutativity, associativity, additive identity, additive inverse, multiplicative identity and distributive properties.

A subset $U$ of $V$ is called a subspace of $V$, if $U$ is also a vector space, with same addition and scalar multiplication as on $V$.

Suppose $U_1,\dots,U_m$ are subsets of $V$. The sum of $U_1,\dots,U_m$, denoted $U_1+\dots+U_m$, is the set of all possible sums of elements.

$$U_1+\dots+U_m=\{ u_1+\dots+u_m : u_i\in U_i \}$$

Suppose $U_1,\dots,U_m$ are subspaces of $V$. Then $U_1+\dots+U_m$ is the smallest subspace of $V$ containing $U_1,\dots,U_m$.

Proof. First of all, $0\in\sum_i U_i$, and $\sum_i U_i$ is closed under addition and scalar multiplication. So $\sum_i U_i$ is a subspace of $V$.

Clearly $U_i$ are contained in $\sum_i U_i$ by making some component $0$. And since subspace must contain all finite sums of their elements, so every subspace of $V$ containing $U_i$ must contain $\sum_i U_i$. Thus, $\sum_i U_i$ is the smallest subspace of $V$ containing $U_i$. $\blacksquare$

The sum is called direct sum, if each element can be written in only one way as a sum.

Suppose $U_i$ are subspaces of $V$. Then $\sum_i U_i$ is a direct sum iff $\sum_i u_i=0$ only has trivial solution $u_i=0$.

Suppose $U,W$ are subspaces of $V$. $U + W$ is direct sum iff $U\cap W=\{ 0 \}$