📂Linear Algebra

직합의 성질

Theorem¹

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

1. $V = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}$
2. It is $V = \sum\limits_{i=1}^{k}W_{i}$, 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. All $v \in V$ can be uniquely expressed in the form of $v = v_{1} + \cdots + v_{k} (v_{i} \in W_{i})$.
4. If $\gamma_{i}$ is an ordered basis of $W_{i}$, then $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is an ordered basis of $V$.
5. There exists an ordered basis $\gamma_{i}$ of $W_{i}$ such that $\gamma_{1} \cup \cdots \cup \gamma_{k}$ becomes an 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 it is $V = \sum\limits_{i=1}^{k}W_{i}$. Let $v_{i} + \cdots v_{k} = 0 (v_{i} \in W_{i})$. Then for some $j$, $$-v_{j} = \sum\limits_{i\ne j}v_{i} \in \sum\limits_{i\ne j}W_{i}$$ Since $v_{j} \in W_{j}$, we obtain the following: $$-v_{j} \in W_{j} \cap \sum\limits_{i\ne j}W_{i} = \left\{ 0 \right\}$$ Therefore, for all $i$, it is $v_{i} = 0$.

• $2. \implies 3.$

  Assume 2. Let it be $v \in V$. Then there exist some $v_{i} \in W_{i}$, which are $v = v_{1} + \cdots + v_{k}$ by assumption. Suppose there exists another expression $v = w_{1} + \cdots + w_{k}\ (w_{i} \in W_{i})$. Then we obtain: $$0 = v - v = (v_{1} - w_{1}) + \cdots + (v_{k} - w_{k})$$ Therefore, it is $v_{i} - w_{i} \in W_{i}$ and since we assumed 2., for all $i$, it is $v_{i} - w_{i} = 0$. Thus, $v = v_{1} + \cdots + v_{k}$ is a unique representation.

• $3. \implies 4.$

  Assume 3. Let $\gamma_{i}$ be an ordered basis of $W_{i}$. By assumption, it is $V = \sum\limits_{i=1}^{k}W_{i}$. This means $\gamma_{1} \cup \cdots \cup \gamma_{k}$ generates $V$. To show this set is linearly independent, assume $v_{ij} \in \gamma_{j}$ and for scalars $a_{ij}$, that $\sum\limits_{i,j} a_{ij} v_{ij} = 0$ is $(j = 1,\dots,m_{i},\ i=1,\dots,k)$. For each $i$, $$w_{i} = \sum_{j=1}^{m_{i}}a_{ij}v_{j}$$ Suppose it holds true for all $i$, then it is $0 = 0 + \cdots + 0 = w_{1} + \cdots + w_{k}$. By assumption, for all $i$, it is $w_{i} = 0$. $$0 = w_{i} = \sum_{j=1}^{m_{i}}a_{ij}v_{j}$$ Since each $\gamma_{i}$ is a basis, it is linearly independent, and therefore for all $i,j$, it is $a_{ij} = 0$. Therefore, $\gamma_{1} \cup \cdots \cup \gamma_{k}$ is linearly independent and is a basis of $V$.

• $4. \implies 5.$

  It is trivial.

• $5. \implies 1.$

  Assume 5. For each $i$, let $\gamma_{i}$ be an ordered basis of $W_{i}$ such that $\gamma_{1} \cup \cdots \cup \gamma_{k}$ becomes an ordered basis of $V$. Then, the sum of the generating sets equals the generating set of the sum, $$\begin{align*} V &= \span(\gamma_{1} \cup \cdots \cup \gamma_{k}) \\ &= \span(\gamma_{1}) + \cdots + \span(\gamma_{k}) \\ &= \sum\limits_{i=1}^{k}W_{i} \end{align*}$$ Now, to prove exclusivity, fix $j (1 \le j \le k)$ and for a non-zero $v \in V$, assume the following: $$v \in W_{j} \cap \sum\limits_{i \ne j} W_{i}$$ By the definition of the intersection, the following holds: $$v \in W_{j} = \span \gamma_{j}, \qquad v \in \sum\limits_{i \ne j} W_{i} = \span(\bigcup_{i \ne j} \gamma_{i})$$ Then $v$ has two different linear combinations in the basis $\gamma_{1} \cup \cdots \cup \gamma_{k}$ of $V$. This contradicts the uniqueness of linear combinations in a basis, thus the assumption is incorrect. Therefore, non-zero elements of $V$ do not belong to $W_{j} \cap \sum\limits_{i \ne j} W_{i}$. $$W_{j} \cap \sum\limits_{i \ne j} W_{i} = \left\{ 0 \right\}$$

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