17 - Subspace

Subspace $U$ of vector space $V$ over $\mathbf{F}$

Non-empty subset of $V$ (so $U\subseteq V$)

• Closed under addition and scalar multiplication

Satisfies the following

1. $U\ne \varnothing$: $U$ is nonempty
   Generally done by showing $0_V\in U$
2. $u_1+u_2\in U$ for all $u_1,u_2\in U$: $U$ is closed under addition
3. $\lambda u\in U$ for all $u\in U$, $\lambda \in \mathbf{F}$: $U$ is closed under scalar multiplication

Addition and Scalar Multiplication are defined the same as $V$

Zero Subspace

$$\{0_V\}$$

Just the $0_V$

Trivial Subspace

$$V$$

Just the vector space itself

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Subspace Test

Let $V$ be a vector space over $\mathbf{F}$ and $U$ a subset of $V$

$U$ is a subspace of $V$ if and only if

1. $0_V\in U$ and

2. $\lambda u_1+u_2\in U$ for all $u_1,u_2\in U$ and $\lambda \in \mathbf{F}$

Proof

Direction $⟹$
Assume $U$ is a subspace of $V$

• $0_V\in U$:
  As $U$ is a subspace then it is non-empty so there exists $u\in U$
  As $U$ is closed under scalar multiplication

$$0u=0_V\in U$$

• $\lambda u_1+u_2\in U$ for all $u_1,u_2\in U$ and all $\lambda \in \mathbf{F}$
  Take $u_1,u_2\in U$ and $\lambda \in \mathbf{F}$
  As $U$ is closed under scalar multiplication then $\lambda u_1\in U$
  As $U$ is closed under addition then $\lambda u_1+u_2\in U$

Direction $⟸$
Assume $0_V\in U$ and that $\lambda u_1+u_2\in U$ for all $u_1,u_2\in U$ and $\lambda \in \mathbf{F}$

• $U$ is non-empty:

$$0_V\in U\text{ by assumption}$$

• $U$ is closed under addition
  Let $\lambda =1$
  For $u_1,u_2\in U$ then $\lambda u_1+u_2=u_1+u_2\in U$

• $U$ is closed under scalar multiplication
  Let $u_2=0_V$
  For $u_1\in U$ and $\lambda \in \mathbf{F}$ then $\lambda u_1+u_2=\lambda u_1+0_V=\lambda u_1\in U$

Hence $U$ is a subspace of $V$

Subspace Notation

$$U\le V⟺U\text{ is a subspace of }V$$

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Properties of Subspaces

Let $V$ be a vector space over $\mathbf{F}$ and let $U\le V$

• $U$ is a vector space over $\mathbf{F}$
  Turns out the only subsets of $V$ that are vector spaces over $\mathbb{F}$ are the subspaces
• If $W\le U$ then $W\le V$
  Subspace of a subspace is also a subspace

Notation on "addition" of subspaces

Let $U,W\le V$ where $V$ is a vector space

$$U+W=\{u+w\mid u\in U,w\in W\}$$

Where $U+W$ is also a subspace of $V$ ($U+W\le V$)

Proof for Subspace

• Condition 1)
  As $0_V\in U$ and $0_V\in W$ then $0_V+0_V=0_V\in U+W$

• Condition2)
  For $v_1,v_2\in U+W$ then by definition of $U+W$ there exists $u_1,u_2\in U$ and $w_1,w_2\in W$ such that

$$v_1=u_1+w_1\text{ and }{v}_2=u_2+w_2$$

As $V$ is a subspace then $\lambda u_1+u_2\in U$
As $W$ is a subspace then $\lambda w_1+w_2\in W$
So

$$(\lambda u_1+u_2)+(\lambda w_1+w_1)=\lambda (u_1+w_1)+(u_2+w_2)=\lambda v_1+v_2\in U+W$$

Hence $U+W$ is a subspace

Notation on "intersection" of subspaces

Let $U,W\le V$ where $V$ is a vector space

$$U\cap W=\{v\mid v\in V\text{ if }v\in U\text{ and }v\in W\}$$

Where $U\cap W$ is also a subspace of $V$ ($U\cap W\le V$)
Largest subspace of $V$ that is contained in both $U$ and $W$