📂Linear Algebra

Properties of Direct Sum

Theorem¹

Let $W_{1}, W_{2}, \dots, W_{k}$ be the subspaces of the finite-dimensional vector space $V$. The following propositions are equivalent.

1. $V = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}$
2. $V = \sum\limits_{i=1}^{k}W_{i}$ and for any vectors $v_{i} \in W_{i}(1 \le i \le k)$, if $v_{1} + \cdots v_{k} = 0$, then for all $i$, $v_{i} = 0$ holds.
3. Every $v \in V$ is uniquely represented in the form of $v = v_{1} + \cdots + v_{k} (v_{i} \in W_{i})$.
4. If $\gamma_{i}$ is the ordered basis for $W_{i}$, then $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is the ordered basis for $V$.
5. There exists an ordered basis $\gamma_{i}$ of $W_{i}$ such that $\gamma_{1} \cup \cdots \cup \gamma_{k}$ becomes the ordered basis of $V$.

Explanation

For two subspaces $W_{1}, W_{2}$ of $V$,

• $W_{1} + W_{2}$ is the sum of $W_{1}, W_{2}$.
• $W_{1} \oplus W_{2}$ is the direct sum of $W_{1}, W_{2}$.

Proof

• $1. \implies 2.$

  Assume 1. Then, $V = \sum\limits_{i=1}^{k}W_{i}$. Let’s say $v_{i} + \cdots v_{k} = 0 (v_{i} \in W_{i})$. Then, for some $j$, $$ -v_{j} = \sum\limits_{i\ne j}v_{i} \in \sum\limits_{i\ne j}W_{i} $$ However, since $v_{j} \in W_{j}$, we get the following. $$ -v_{j} \in W_{j} \cap \sum\limits_{i\ne j}W_{i} = \left\{ 0 \right\} $$ Therefore, for all $i$, $v_{i} = 0$ holds.

• $2. \implies 3.$

  Assume 2. Let’s say $v \in V$. Then, by assumption, there exist $v_{i} \in W_{i}$ such that $v = v_{1} + \cdots + v_{k}$. Suppose another expression $v = w_{1} + \cdots + w_{k}\ (w_{i} \in W_{i})$ exists. Then, we obtain the following. $$ 0 = v - v = (v_{1} - w_{1}) + \cdots + (v_{k} - w_{k}) $$ Therefore, $v_{i} - w_{i} \in W_{i}$, and by assuming 2., for all $i$, $v_{i} - w_{i} = 0$ holds. Hence, $v = v_{1} + \cdots + v_{k}$ is a unique expression.

• $3. \implies 4.$

  Assume 3. Let $\gamma_{i}$ be the ordered basis for $W_{i}$. By assumption, $V = \sum\limits_{i=1}^{k}W_{i}$ holds. This means that $\gamma_{1} \cup \cdots \cup \gamma_{k}$ spans $V$. Now, to show this set is linearly independent, let’s say $v_{ij} \in \gamma_{j}$, and for scalars $a_{ij}$, $\sum\limits_{i,j} a_{ij} v_{ij} = 0$ $(j = 1,\dots,m_{i},\ i=1,\dots,k)$. For each $i$, $$ w_{i} = \sum_{j=1}^{m_{i}}a_{ij}v_{j} $$ Since $0 \in W_{i}$ for all $i$, $0 = 0 + \cdots + 0 = w_{1} + \cdots + w_{k}$ holds. Then, by assumption, for all $i$, $w_{i} = 0$ holds. $$ 0 = w_{i} = \sum_{j=1}^{m_{i}}a_{ij}v_{j} $$ But each $\gamma_{i}$ is a basis and thus linearly independent. Therefore, for all $i,j$, $a_{ij} = 0$ holds. Hence, $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is linearly independent and is a basis for $V$.

• $4. \implies 5.$

  Trivial.

• $5. \implies 1.$

  Assume 5. For each $i$, let’s consider $\gamma_{i}$ to be the ordered basis of $W_{i}$ that makes $\gamma_{1} \cup \cdots \cup \gamma_{k}$ the ordered basis of $V$. Then, since the sum of unions and the union of sums are the same, $$ \begin{align*} V &= \span(\gamma_{1} \cup \cdots \cup \gamma_{k}) \\ &= \span(\gamma_{1}) + \cdots + \span(\gamma_{k}) \\ &= \sum\limits_{i=1}^{k}W_{i} \end{align*} $$

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