[Linear Algebra] Subspace

Subspace

$\mathbf{Definition.}$ Subspace
Let $V$ be a vector space over $F$. A subset $W$ of $V$ is called a subspace of $V$ if $W$ is a vector space under the operations (addition & scalar multiplication) defined in $V._\blacksquare$

In other words, we need to confirm that $W$ must satisfy the various properties of vector space to define subspace. However, there is a lemma that can make this process simpler.

$\mathbf{Lemma.}$ Let $V$ be a vector space over $F$.
A non-empty subset $W$ of $V$ is a subspace if and only if $c \cdot x + y \in W$ for $\forall x, y \in W, \forall c \in F._\blacksquare$

$\mathbf{Proof.}$

( $\Rightarrow$ ) $c \cdot x \in W$, $y \in W \to c \cdot x + y \in W$

( $\Leftarrow$ ) $-1 \cdot x + x = \mathbf{0} \in W, -x = -1 \cdot x + \mathbf{0} \in W$.
$1 \cdot x + y = x + y \in W$, $c \cdot x + \mathbf{0} = c \cdot x \in W$. So it is closed to addition and scalar multiplication. Since $W$ is a subset of $V$, other properties satisfy automatically.

Here are some examples.

$\mathbf{Example.}$ Examples of subspace

1. { $\mathbf{0}$ } of $V$ is a trivial subspace of $V$.

2. { $(0, a_2, \cdots, a_n) | a_i \in F$ } is a subspace of $F^n$.

3. Let $W$ be a set of all symmetric matrices in $M_{n \times n} (F)$. Then, $\forall A, B \in W \to c \cdot A + B$ is also symmetry. By the previous lemma, $W$ is a subspace of $M_{n \times n} (F)$.

4. Let $W$ be a set of all continuous functions $f: [a, b] \in R$. Then, $W$ is a subspace of the vector space of all functions $f: [a, b] \in R$ as the scalar multiple of continuous functions and the addition of two continuous functions are also continuous.

5. Let $W$ be a set of all polynomials of $\text{deg } \leq n$. Then, $W$ is a subspace of $F[x]$ as $\text{deg }(c \cdot f + g) \leq n$.

Let $U, W$ be subspaces of $V$.

6. $U + W :=$ { $u + w | u \in U, w \in W$ } is a subspace of $V$:
$\forall u_1, u_2 \in U$ and $\forall w_1, w_2 \in W$, $u_1 + w_1, u_2 + w_2 \in U + W$ and $(c \cdot u_1 + u_2) + (c \cdot w_1 + w_2) \in U+W._\blacksquare$

7. $U \cap W$ is a subspace of $V$:
Let $\forall \alpha, \beta \in U \cap W$, $c \in F$ be given. Then, $c \cdot \alpha + \beta \in V$ since $\alpha \in U, \beta \in U$ and $U$ is a subspace $._\blacksquare$

8. But, $U \cup W$ may not be a subspace of $V$. Consider $V = \mathbb{R}^2$ and $U$: $x$-axis, $W$: $y$-axis.
Then it is clear that $(1, 1) \notin U \cup W$.

9. If $U \subset V_1$, $W \subset V_2$ are subspaces, then $U \times W$ is a subspace of $V_1 \times V_2$:
$\forall (u_1, w), (u_2, w)$, $\forall c \in F$, $c \cdot (u_1, w) + (u_2, w) \in U \times W._\blacksquare$

10. Let $W =$ { $(c_1, c_2, c_3) \in F^3 | 2c_1 + c_2 - 5c_3 = 0$ }. Then it is a subspace of $F^3$.
But $W =$ { $(c_1, c_2, c_3) \in F^3 | 2c_1 + c_2 - 5c_3 = 1$ } is not a subspace of $F^3$ as $\mathbf{0} \notin W$.

Span

$\mathbf{Definition.}$ Span
Let $s$ be a nonempty subset of vector space $V$ over $F$. The span of $S$ is defined as follows:
$\text{Span}(S) := \underset{S \subset W \subset V}{\bigcap} W$, where $W$ is a subspace of $V$.

The following statements are useful lemmas to understand span and to prove the further theorem.

$\mathbf{Lemma.}$
1. $S \subset T \subset V \to \text{Span}(S) \subset \text{Span}(T)$
2. If $S$ is a subspace of $V$, then $\text{Span}(S) = S$
3. $\text{Span}(S) =$ { $c_1 \cdot s_1 + \cdots + c_n \cdot s_n | c_i \in F, s_i \in S, n \geq 1$ }.

$\mathbf{Proof.}$

1. Let $w \in S \subset \text{Span}(S)$. Then, $w \in S \subset T \in \underset{T \subset W \subset V}{\bigcap} W$.

2. ( $S \subset \text{Span}(S)$ ) Clear.

( $\text{Span}(S) \subset S$ ) Since $S$ is a subspace of $V$, $S \in$ { $W | S \subset W \subset, W \text{ is a subspace }$ }.

3. For convenience, let $T =$ { $c_1 \cdot s_1 + \cdots + c_n \cdot s_n | c_i \in F, s_i \in S, n \geq 1$ }.
First, $T \subset \text{Span(S)}$ as $\text{Span(S)}$ is a subspace: if $\alpha \in T$, $\alpha$ is a linear combination of elements of $\text{Span(S)}$.
Conversely, if $x, y \in T$, then $c \cdot x + y \in T$ is a linear combination, thus $T$ is a subspace containing $S$. Thus, $\text{Span(S)} \subset T._\blacksquare$

$\mathbf{Remark.}$
The span of $S$ is also a subspace since it is defined as a intersection of subspaces.

Here are some examples.

$\mathbf{Example.}$ Examples of subspace

1. $\mathbb{R}^n = \text{Span}(e_1, \cdots, e_n)$

2. The vector space of polynomials of $\text{deg} \leq n$ is $\text{Span}(1, x, \cdots, x^n)$.

3. $M_{n \times n} (F) = \text{Span}$ { $E_{ij} | 1 \leq i, j \leq n$ }.

Reference

[1] K. Hoffman, and R. Kunze. Linear Algebra PHI Learning, Second edition, 2004