Existence of the Direct Sum of Vector Spaces 📂Linear Algebra

Existence of the Direct Sum of Vector Spaces

Theorem

Let a vector space $V$ of dimension $n \ge 2$ and a subspace $W \lneq V$ of $V$ be given. Then there exist $W$ and another subspace $U$ such that the following holds.

$$ V = W \oplus U $$

In other words, $V$ can be represented as a direct sum of a subspace and another distinct subspace.

Explanation

The $U$ in the theorem is also called a complementary subspace of $W$. The complementary subspace of $W$ is not unique. For example, suppose $V = \mathbb{R}^{2}$ and $W = \span \left\{ (1,0) \right\}$. Then $W_{1} = \span \left\{ (0,1) \right\}$ and $W_{2} = \span \left\{ (1,1) \right\}$ are both complementary subspaces of $W$. That is, $\mathbb{R}^{2}$ can be written as follows.

$$ \mathbb{R}^{2} = \span \left\{ (1,0 ) \right\} \oplus \span \left\{ (0,1) \right\} = \span \left\{ (1,0) \right\} \oplus \span \left\{ (1,1) \right\} $$

The reason the dimension in the theorem is set to $n \ge 2$ is that when $n = 1$, only the trivial direct sum ▷eq19◯ exists.

Proof

Let $V$ be a $n$-dimensional vector space and let $W$ be a $k$-dimensional subspace of $V$. Let the basis of $W$ be $\left\{ w_{1}, w_{2}, \ldots, w_{k} \right\}$. Then by the following lemma we can obtain a basis $\beta = \left\{ w_{1}, w_{2}, \dots, w_{k}, u_{1}, u_{2}, \dots, u_{n-k} \right\}$ of $V$.

Basis extension theorem
Let $W \le V$ be a subspace of the $n$-dimensional vector space $V$. Let $\gamma = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k} \right\}$ be a basis of $W$. Then by adding appropriate elements to $\gamma$ one can extend it to a basis $\beta = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k}, \mathbf{v}_{k+1}, \dots, \mathbf{v}_{n} \right\}$ of $V$.

Since $u_{1}, \dots, u_{n-k}$ is linearly independent, $\left\{ u_{i} \right\}$ is a basis of the space $U$ generated by them.

$$ U = \span \left\{ u_{1}, u_{2}, \dots, u_{k} \right\} $$

Now we show that such $U$ is a complementary subspace of $W$ and satisfies $V = W \oplus U$.

Condition for $V = W \oplus U$
For any $v = w + u$ there exist $w \in W$ and $u \in U$ satisfying $v = w + u$.
$W \cap U = \left\{ 0 \right\}$

Since the bases of $V$, $W$, and $U$ were chosen as above, the following holds. $$ \begin{align*} v &= a_{1}w_{1} + a_{2}w_{2} + \cdots a_{k}w_{k} + b_{1}u_{1} + b_{2}u_{2} + \cdots b_{n-k}u_{n-k} \\ &= \sum_{i} a_{i}w_{i} + \sum_{j}b_{j}u_{j} \end{align*} $$
Given $W = \span \left\{ w_{1}, \dots, w_{k} \right\}$ and $U = \span \left\{ u_{1}, \dots, u_{n-k} \right\}$, and since $\beta$ is a basis of $V$ (hence linearly independent), $W \cap U = \left\{ 0 \right\}$ follows.

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