direct sum

def (sum of subspace)

Let $u_{1} \dots u_{m}$ be subspaces of $V$ . The sum is defined as

u_{1} + \dots + u_{m} = {u_{1} + \dots + u_{m} | u_{i} \in U_{i} i \in [1, m]}

i.e. the collection of all possible sums of elements $u_{1} \dots u_{m}$

Ex: Consider $V = R^{2}$ . Let $U_{1} = {(x, 0) | x \in R}$ and $U_{2} = {(0, y) | y \in R}$

\begin{aligned} U_{1} + U_{2} & = {(x, 0) + (0, y) | x, y \in R} \\ = {(x, y) | x, y \in R} = R^{2} \end{aligned}

Proposition: Let

U_{1} \dots U_{m}

be subspaces of

V

. Then

U_{1} + \dots + U_{m}

is the smallest subspace of

V

containing

U_{1} \dots U_{m}

Proof:

Sum is a subspace: Let $S = U_{1} + \dots + U_{m}$ . We need to prove $0 \in S$ . Since each $U_{i}$ is a subspace, there is a $0$ in each subspace.

0 = 0 (\in U_{1}) + 0 (\in U_{2}) + \dots + 0 (U_{m}) \in S

Now we need to show $S$ is closed under addition. Let $u, v \in S$

\begin{aligned} u & = u_{1} + \dots + u_{m} \\ v & = v_{1} + \dots + v_{m} \\ u + v & = (u_{1} + v_{1}) + \dots + (u_{m} + v_{m}) \in S \end{aligned}

Sum is the smallest subspace: Consider $u_{i} \in U_{i}$ . We can express this as: $u_{i} = 0 + 0 + \dots + u_{i} + 0 + \dots + 0 = 0 \in S$ Thus each $U_{i} \subseteq S ⟹ S$ contains each $u_{i}$ . Consider $T$ is a subspace containing $U_{1} \dots U_{m}$ . Since T is a subspace it is closed under addition and therefore contains all sums of elements from $u_{1} \dots U_{m}$ and $S \subseteq T$

def (Direct Sum)

Suppose $U_{1} \dots U_{m}$ are subspaces of $V$ . Then the sum $U_{1} \dots U_{m}$ is called a direct sum if each $v \in u_{1} + \dots + u_{m}$ can be written in a unique way as $v = u_{1} + \dots + u_{m}$ where $u_{i} \in U_{i}$ . In this case the notation is $U_{1} \oplus \dots \oplus U_{m}$

Theorem: Suppose $U_{1} \dots U_{m}$ are subspaces of $V$ . Then $U_{1} + \dots + U_{m}$ is a direct sum iff the only way to write $0$ as a sum of the form $u_{1} + \dots + u_{m}$ is by taking each $u_{i} = 0$

Proof:

[=>] If $u_{1} + \dots + u_{m}$ is a direct sum, since $0 \in u_{1} + \dots + u_{m}$ there is only way to write $0 = u_{1} + \dots + u_{m}$ (by defn.)

[<=] Suppose the unique way to write 0 is $0 = 0 + 0 + \dots + 0$ . Suppose $v \in U_{1} + \dots + U_{m}$ and $v_{i}, w_{i} \in U_{i}$

\begin{aligned} v & = v_{1} + \dots + v_{m} \\ v & = u_{1} + \dots + u_{m} \\ v - v & = (v_{1} - u_{1}) + \dots + (v_{m} - u_{m}) \\ 0 & = (v_{1} - u_{1}) + \dots + (v_{m} - u_{m}) \end{aligned}

By our hypothesis: $v_{i} - u_{i} = 0 ⟹ v_{i} = u_{i}$ This is a contradiction.

Proposition: If

U

and

W

are subspaces of

V

, then

U + W

is a direct sum iff

U \cap W = {0}

Proof:

[=>] Suppose $U + W$ is a direct sum. Let $v \in U \cap W$ . Then

\begin{aligned} v & = v (\in U) + 0 (\in W) \\ = 0 (\in U) + v (\in W) \end{aligned}

These must be the same since it is a direct sum. Thus $v = 0 ⟹ U \cap W = {0}$

[<=] Suppose $U \cap W = {0}$ . By the previous theorem it is sufficient to check if there is a unique way to write $0$ as a sum of elements of $U, W$ . Suppose if $u \in U, w \in W$

\begin{aligned} 0 & = u + w \\ u (\in U) & = - w (\in W) \in U \cap W \end{aligned}

Which is by our hypothesis i.e. $0 = u + w ⟹ u = w = 0$ .