# Direct Sum Theorems

Recall the following definition and lemma regarding the direct sum of a set of subspaces $U_1, U_2, …, U_m$ of a vector space $V$.

Definition: Let $U_1, U_2, …, U_m$ all be vector subspaces of the $\mathbb{F}$-vector space $V$. We define the direct sum of these subspaces $\bigoplus_{i=1}^{m} = U_1 \oplus U_2 \oplus … \oplus U_m = V$ is defined to be a sum of the subspaces $U_1, U_2, …, U_m$ to which each element in $V$ can be uniquely written as $u_1 + u_2 + … + u_m$ where $u_i \in U_i$ for $i = 1, 2, …, m$. |

Lemma 1: Let $U_1, U_2, …, U_m$ be vector subspaces of the $\mathbb{F}$-vector space $V$. Then these subspaces form a direct sum $\bigoplus_{i=1}^{m} U_i = V$ if and only if the sum of these subspaces is equal to $V$, that is $\sum_{i=1}^{m} U_i = V$ and when $\sum_{i=1}^{m} u_i = 0$ where $u_i \in U_i$ implies that $u_i = 0$ for every $i = 1, 2, …, m$. |

We will now look at some more important theorems regarding the direct sum of subspaces.

Theorem 1: If $U_1$ and $U_2$ are subspaces of the vector space $V$ over the field $\mathbb{F}$, then $U_1 + U_2 = U_1 \oplus U_2 = V$ if and only $U_1 \cap U_2 = \{ 0 \}$. |

Intuitively this should make sense. The subspaces $U_1$ and $U_2$ can only form a direct sum of $V$, that is $V = U_1 \oplus U_2$ if these subspaces share no vectors in common apart from the zero vector. For example, if $U_1$ contains a nonzero vector from $U_2$, call it $x$, that is $x \in U_1 \cap U_2$, then:

(1)

\begin{align} \underbrace{x}_{\in U_1} + \underbrace{0}_{\in U_2} = \underbrace{0}_{\in U_1} + \underbrace{x}_{\in U_2} \end{align}

Therefore not all sums of vectors $u_1 + u_2$ are uniquely determined since $x$ can be derived in two different ways. We will now demonstrate the formal proof.

**Proof:**$\Rightarrow$ Suppose that $U_1 + U_2 = U_1 \oplus U_2 = V$. We first want to show that $U \cap U_2 = \{ 0 \}$. We note that since $V = U_1 \oplus U_2$, then if $u_1 + u_2 = 0$ implies that $u_1 = 0$ and $u_2 = 0$.

- Now $0 = x + (-x)$, so $x \in U_1$ and $-x \in U_2$, and since $V = U_1 \oplus U_2$ this implies that $x = -x = 0$.

- $\Leftarrow$ Suppose that $U_1 \cap U_2 = \{ 0 \}$. We want to show that $U_1 + U_2 = U_1 \oplus U_2 = V$. Now suppose that for $u_1 \in U_1$ and $u_2 \in U_2$ that $0 = u_1 + u_2$. Therefore $u_1 = -u_2$. Therefore $u_1 = -u_2 \in U_1 \cap U_2 = \{ 0 \}$ and so $u_1 = -u_2 = 0$ or rather $u_1 = u_2 = 0$ and thus by lemma 1 since $\sum_{i=1}^{m} u_i = 0$ implies that $u_1 = u_2 = 0$, we have that $U_1 + U_2 = U_1 \oplus U_2 = V$ $\blacksquare$

Corollary 1: If $U_1, U_2, …, U_m$ are subspaces of the vector space $V$ over the field $\mathbb{F}$, then $\sum_{i=1}^{m} U_i = \bigoplus_{i=1}^{m} U_i = V$ if and only $U_i \cap \left ( \sum_{j \neq i} U_j \right) = \{ 0 \}$ where $i = 1, 2, …, m$. |

**Proof:**$\Rightarrow$ Suppose that $\sum_{i=1}^{m} U_i = \bigoplus_{i=1}^{m} U_i =V$, that is the sum is direct. We first want to show that $U_i \cap \left ( \sum_{j \neq i} U_j \right) = \{ 0 \}$ where $i = 1, 2, …, m$. Now since $V = \oplus_{i=1}^{m} U_i$, then this implies that if $u_1 + u_2 + … u_m = 0$ implies $u_1 = u_2 = … = u_m$. Without loss of generality consider $U_1 \cap \left ( \sum_{j=2}^{m} U_j \right)$. Then $u_1 \in U_1$ and $(u_2 + u_3 + … + u_m) \in \sum_{j=2}^{m} U_j$.

- Now if $0 = x + (-x)$ then $x \in U_1$ and $-x \in \sum_{j=2}^{m} U_j$ and since $V = U_1 \bigoplus_{j=2}^{m} U_i$ this implies that $x = -x = 0$.

- $\Leftarrow$ Now suppose that $U_i \cap \left ( \sum_{j\neq i}^{m} U_j \right ) = \{ 0 \}$. We want to show that $\sum_{i=1}^{m} U_i = \bigoplus_{i=1}^{m} U_i = V$. Without loss of generality let $U_1 \cap \left ( \sum_{j=2}^{m} U_j \right ) = \{ 0 \}$, and let $u_1 \in U_1$ and $u_2 \in \left ( \sum_{j=2}^{m} U_j \right )$ such that $u_1 + u_2 = 0$. Therefore $u_1 = -u_2$ and so $u_1 = -u_2 \in U_1 \cap \left ( \sum_{j=2}^{m} U_j \right ) = \{ 0 \}$ which implies that $u_1 = -u_2 = 0$ or rather $u_1 = u_2 = 0$, and so we have that $\sum_{i=1}^{m} U_i = \bigoplus_{i=1}^{m} U_i = V$. $\blacksquare$