Skip to content

10.28 Subspaces

Subspaces

Subspaces

Definition

Let \(V\) be an \(\mathbb{F}\)-vector space, a subspace \(W\) of \(V\) is a subset of \(V\) (\(W\subseteq V\)) such that with the restrictions of the sum and the multiplication by scalars, it is an \(\mathbb{F}\)-vector space.

Remark

It is implicitly assumed that: \(+|_{W\times W}: W\times W\rightarrow W\) and \(\cdot|_{\mathbb{F}\times W}:\mathbb{F}\times W\rightarrow W\)

If \(\tilde 0\) is a neutral element of \(W\), thus \(\tilde 0=0\)

For any \(w\in W,\exists \tilde w\) such that \(w+\tilde w=0,\tilde w=-w\)

Proposition

A non-empty subset \(W\) of \(V\) is a subspace of \(V\) iff

  • \(\forall v,w\in W,v+w\in W\)​
  • \(\forall \lambda \in\mathbb{F},\forall v\in W,\lambda\cdot v\in W\)​

Proof

\(\Rightarrow\)) Direct from the definition

\(\Leftarrow\)) Let's prove that \((W,+,\cdot )\) is a \(\mathbb{F}\)-vector space

  1. \(+\) is associative and commutative by particularization (in large space yes,you can also in small space)

  2. \(\cdot\) is distributive both by particularization 在v里可以 w自然可以

  3. \(\cdot\) is associative by particularization

  4. \(\exists\) of neutral element: let \(w\in W\), then \(0=0\cdot w\in W\) and it is a neutral element for \(W\) by particularization

  5. Given \(w\in W,\exists\) on opposite \(-w=(-1)\cdot w\in W\) and it is an opposite of \(w\in W\) by particularization

Examples

  1. \(W=\{0\},W=V\) are the trivial subspaces of \(V\)​

  2. \(V=\R^3\), \(W_1=\{(x,y,0):x,y\in\mathbb{R}\}\subsetneq V\)

    \((x,y,0)+(a,b,0)=(x+a,y+b,0)\in W_1\)

    \(l(x,y,0)=(lx,ly,0)\in W_1\)

    After check the proposition above, Yes

    \(W_2=\{(x,y,y):x,y\in\mathbb{R}\}\) Yes

    \(W_3=\{(x,2x,\sqrt2x):x\in\R\}\) Yes

    \(W_4=\{(x,x+z,z):x,z\in\R\}\) Yes

  3. \(V=\R^3\) \(W_1=\{(x,y,1):x,y\in \R\}\) No!

    \(W_2=\{(x,\sqrt{\left|x\right|},0):x\in\mathbb{R}\}\) No!

Question

  1. Which are all the subspaces of \(V=\R^2\)

    • if \(v\neq 0,v\in W\Rightarrow \lambda\cdot v\in W\)

    Moreover \(W=\{\lambda v:\lambda\in\R\}\) is a subspace * If \(W\) contains 2-non-colinear vectors \(v\) and \(w\), hence a goof 2-list

    So, any \(\forall z\in\R\), \(z=\lambda\cdot v+\mu\cdot w\in W\)

    Conclusion: \(W\) is a non-trivial subspaces of \(\R^2\) iff \(W\) is a line through 0

    image​

  2. Which are all the subspaces of \(V=\R^3\)​

    • The zero subspace \(\{0\}\),
    • All lines through the origin,
    • All planes through the origin,
    • The full space \(\mathbb{R}^3\).

Sum of subspaces

Definition

Given subspaces \(W_1,W_2,W_3,...,W_{k}\) of \(V_1\)

The sum of them is the subspace \(W_1+...+W_{k}=\{w_1+\cdots+w_{k}:w_{i}\in W_{i}\}\)

Remark

\(W_1+...+W_{k}\) is in fact a subspace of \(V\)

  1. \((w_1+w_2+...+w_{k})+(\widetilde{w}_1+\widetilde{w}_2+...+\widetilde{w}_{k})=(w_1+\widetilde{w_1})+...+(w_{k}+\widetilde{w_{k}})\in W_1+...+W_{k}\)​

  2. \(\lambda(w_1+w_2+...+w_{k})=\lambda w_1+...+\lambda w_{k}\in W_1+...+W_{k}\)

Example

\(V=\R^3,W_1=\{(x,0,0):x\in\R\},W_2=\{(0,y,0):y\in\R\}\Rightarrow W_1+W_2=\{(x,y,0):x,y\in\R\}\)​

image​

\(V=\mathbb{C}^3,W_1=\{(x,y,x+y):x,y\in\mathbb{C}\},W_2=\{(x,x+z,z):x,z\in\mathbb{C}\}\Rightarrow W_1+W_2=\mathbb{C}^3\)

Since \((1,0,1)\in W_1\) \((0,1,1)\in W_1\) \((1,1,0)\in W_2\) \((0,1,1)\in W_2\)

\(W_1\cap W_2=?~~~~~~~\) \((x,y,x+y)=(a,a+b,b)\Leftrightarrow x=a,y=a+b,x+y=b\Rightarrow x=0,y=b\) \(W_1\cap W_2=(0,y,y)~\)​

Proposition

The intersection of two subspaces of \(V\) is a subspace. Moreover, the intersection of an arbitrary family of subspaces is a subspace

Proof

Let \(W_1\) and \(W_2\) subspaces. Consider \(W_1\cap W_2\)

  • \(v,w\in W_1\cap W_2\Rightarrow v+w\in W_1\) and \(v+w\in W_2\Rightarrow v+w\in W_1\cap W_2\)
  • \(v\in W_1\cap W_2\Rightarrow \lambda \cdot v\in W_1\) and \(\lambda \cdot v\in W_2\Rightarrow \lambda\cdot v\in W_1\cap W_2\)​

Now, given \(\{W_{i}\}_{i\in I}\) a family of subspaces, \(\bigcap_{i\in I}W_{i}\) is a subspace

  • \(v,w\in\bigcap_{i\in I}W_{i}\Rightarrow v+w\in W_{i},\forall i\in I\) \(\Rightarrow v+w\in\bigcap_{i\in I}W_{i}\)
  • \(v\in\bigcap_{i\in I}W_{i}\Rightarrow\lambda\cdot v\in W_{i},\forall i\in I\) \(\Rightarrow\lambda\cdot v\in\bigcap_{i\in I}W_{i}\)

Is \(W_1\cup W_2\) a subspace? No in general

Because \(W_1\cup W_2\) is just two lines, will not be a plane

image​

Proposition

Given \(W_1,W_2,...,W_{k}\) subspaces of \(V\), the sum \(W_1+W_2+...+W_{k}\) is the smallest subspace of \(V\) containing \(W_1,W_2,...,W_{k}\)

Proof \(W_1+...+W_{k}\) contains \(W_1,W_2,...,W_{k}\) (just let other take \(0\in W_i\))

Let \(W\) be another subspace also containing \(W_1,W_2,...,W_{k}\)

Given \(w_1\in W_1,w_2\in W_2,...,w_{k}\in W_{k}\Rightarrow w_1+w_2+...+w_{k}\in W\). Hence \(W_1+W_2+...+W_{k}\subseteq W\)​

Definition

Given a list of vectors \(v_1,v_2,...,v_k\) of \(V\), the subspace generated by them is the subspace \(W_1+W_2+...+W_{k}\) where \(W_i=\{\lambda\cdot v_i,\lambda\in\mathbb{F}\}\)​

Notation: \(\langle v_1,v_2,...,v_k\rangle\)

Remark: \(\langle v_1,v_2,...,v_{k}\rangle=\{\text{all linear combinations of }v_1,v_2,...,v_{k}\}=\{\lambda_1v_1+\lambda_2v_2+\cdots+\lambda_{k}v_{k}:\lambda_{i}\in\mathbb{F},i=1,\ldots,k\}\)

Example

\(V=\R^2\) \(W_1=\langle\left(1,1\right),\left(0,1\right)\rangle=\)

\((a,b)=a(1,1)+(b-a)(0,1)\)

image​

\(W_2=\langle(1,-1),(1,1),(1,2)\rangle=\mathbb{R}^2\)​

‍