12.2 Product spaces

Let \(V_1,...,V_n\) be finite-dimensional vector spaces

Definition

Product spaces

The product of \(V_1,...,V_k\) is the vector space \(V=V_1\times...\times V_k\) (Cartesian product)with \(+\) and \(\cdot\) defined by \((v_1,...,v_k)+(v_1',...,v_k')=(v_1+v_1',...,v_k+v_k')\) and \(\lambda(v_1,...,v_k)=(\lambda v_1,...,\lambda v_k)\)

Notation

If \(V_{i}=V,\forall i=1,...,k\), then we write \(V^{k}=V\times...\times V\)

• \(\mathbb{R}^{n}=\mathbb{R}\times\ldots\times\mathbb{R}\)​
• \(\mathbb{F}^{n}=\mathbb{F}\times\ldots\times\mathbb{F}\)

Example

\(V_1=\mathbb{C}^2,V_2=M_{2\times2}\left(\mathbb{C}\right),V_3=\mathbb{C}_1\left\lbrack x\right\rbrack\)

\(V=V_1\times V_2\times V_3\), then \(v=\left(\left(1,i\right),\begin{pmatrix}i & 0\\ 1 & 2\end{pmatrix},ix+2\right)\)

Projection and inclusion

Now \(V_1\times...\times V_{k}\xrightarrow{\pi_{i}}V_{i}\), the i-th projection

\((v_1,...,v_{k})\mapsto v_{i}\) \(\pi_i\) is linear

\(V_{i}\xrightarrow{inc_i}V_1\times...\times V_k\) the i-th inclusion

\(v_{i}\mapsto\left(0,\ldots,v_{i},\ldots,0\right)\) \(inc_i\) is linear

Q: Are they inverse of each other?

1. \(V_{i}\xrightarrow{inc_i}V_1\times...\times V_k\xrightarrow{\pi_i}V_i\)​

   \(v_{i}\mapsto\left(0,\ldots,v_{i},\ldots,0\right)\mapsto v_{i}\)​

   Then \(\pi\circ inc_{i}=Id_{v_i}\)​

2. \(V_1\times...\times V_{k}\xrightarrow{\pi_{i}}V_{i}\xrightarrow{inc_{i}}V_1\times...\times V_{k}\)

   \(\left(v_1,\ldots,v_{k}\right)\mapsto v_{i}\mapsto\left(0,\ldots,v_{i},\ldots,0\right)\)​

   Then \(inc_{i}\circ\pi\neq Id_{v_1\times...\times v_k}\)

   In general

Remark

\(inc_{i}:V_{i}\rightarrow V_1\times\ldots\times V_{k}\) is injective. Hence \(V_{i}\simeq range\left(inc_i\right)=0\times...\times V_i\times...\times0=V_i'\)

Although direct sum is about subspace, but we restrict the codomain to the range, then it is under the subspaces

Claim: \(V_1\times...\times V_{k}\simeq V_1^{\prime}\oplus\ldots\oplus V_{k}^{\prime}\) \([\left(v_1,\ldots,v_{k}\right)\longmapsto inc_{i}\left(v_1\right)+\cdots+inc_{k}\left(v_{n}\right)]\)​

Consequence: \(\dim\left(V_1\times\ldots\times V_{k}\right)=\dim V_1^{\prime}+\cdots+\dim V_{k}^{\prime}=\dim V_1+\cdots+\dim V_{k}\)

From basis of \(V_1,...,V_k\) one gets a basis of \(V_1\times ...\times V_k\)

Example

V=V_1\times V_2\times V_3, then v=\left(\left(1,i\right),\begin{pmatrix}i & 0\ 1 & 2\en...

Basis: \(\left(\left(1,0\right),0,0\right),\left(\left(0,1\right),0,0\right),\left(0,\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix},0,0\right),\left(0,\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},0,0\right),\\\left(0,\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},0,0\right),\left(0,\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix},0,0\right),\left(0,0,1\right),\left(0,0,x\right)\)​

8 bases

Proposition

There are canoncial isomorphisms

1. \(L(V_1\times...\times V_{k},W)\simeq L(V_1,W)\times\ldots\times L(V_{k},W)\)​

2. \(L(V,W_1\times...\times W_{k})\simeq L(V,W_1)\times\ldots\times L(V,W_{k})\)

Proof 1 Check their dimension are same, then by theorem there exists a isomorphism but it must be finite dimension

Proof 2

1. Define \(\Phi:L(V_1\times...\times V_{k},W)\rightarrow L(V_1,W)\times\ldots\times L(V_{k},W)\)​

   \(f\mapsto\left(f_1,\ldots,f_{i},\ldots,f_{k}\right)\) where \(f_i=f\circ inc_i\)

   \(f:(v_1,....,v_k)\mapsto w\) and \(inc_{i}:v_{i}\mapsto\left(0,\ldots,v_{i},\ldots,0\right)\)

   Then \(f_{i}=f\circ inc_{i}=f\left(0,\ldots,v_{i},\ldots,0\right)=w_{i}\), thus \(f_i\in L(V_i,W)\)

   Take \(\Phi\left(f+\lambda g\right)=\left(f_1+\lambda g_1,\ldots,f_{i}+\lambda g_{i},\ldots,f_{k}+\lambda g_{k}\right)=\Phi\left(f\right)+\lambda\Phi\left(g\right)\) linear

   Injective: \(\Phi\left(f\right)=0\Rightarrow\left(f_1,\ldots,f_{i},\ldots,f_{k}\right)=0\Rightarrow f_{i}=0,\forall i\Rightarrow f\circ inc_{i}=0,\forall i\)

   Then \(f(inc_{i})=0\Rightarrow f\left(\left(0,\ldots,v_{i},\ldots,0\right)\right)=0\Rightarrow w_{i}=0\)​

   Then \(f=0\Rightarrow\text{Null}\Phi=0\)​

   Surjective: Let's take any \(\left(f_1,\ldots,f_{i},\ldots,f_{k}\right)\), then let \(f=f_{i}\circ\pi_{i}\)

   Then \(\Phi\left(f\right)=\left(f\circ inc_{1},\ldots,f\circ inc_{i},\ldots,f\circ inc_{k} \right)=\left(f_{1}\circ\pi_{1}\circ inc_{1},\ldots,f_{i}\circ\pi_{i}\circ inc_{i} ,\ldots,f_{k}\circ\pi_{k}\circ inc_{k}\right)=\left(f_{1},\ldots,f_{i},\ldots,f_{k} \right)\)

   Thus it is a isomorphism

2. Here

Quotient spaces

Given \(U\hookrightarrow V\), we'll define \(V/U\)

For \(v\in V\), \(v+U=\{v+u:u\in U\}\) is the affine subspace of \(V\) parallel to \(U\) through \(v\)

Example

\(V=\R^2,U=/\)​

​

Remark

\(w_1,w_2\in v+U\Rightarrow w_1-w_2\in U\)

Proof

\(w_1=v+u_1,w_2=v+u_2\Rightarrow w_1-w_2=u_1-u_2\in U\)

The relation defined on \(V\) by \(v_1\sim v_2\) iff \(v_1-v_2\in U\) is an equivalence relation (check: \(v_1\sim v_1,v_1\sim v_2\Rightarrow v_2\sim v_1,v_1\sim v_2\wedge v_2\sim v_3\Rightarrow v_1-v_2+v_2-v_3=v_1-v_3\in U\Rightarrow v_1\sim v_3\))

Notation

\([v]=v+U\)

Recall

The set of equivalence classes is the "quotient set \(X/\sim\)" (\(\sim\) is the notation of equivalence relation)

In this case we denote it by: \(V/U=\left\lbrace\left\lbrack v\right\rbrack:v\in V\right\rbrace\)

equivalence classes are disjoint or they coincide: \((v+U)\cap(w+U)\neq\emptyset\;\iff\;v+U=w+U\)

Lemma

The following are equivalent

\(v-w\in U\)/ \(v+U=w+U\)/ \((v+U)\cap(w+U)\neq\emptyset\;\)

Now, let's define \(+\) and \(\cdot\) on \(V/U\)

\([v]+[w]=[v+w]\) \(\left(\left(v+U\right)+\left(w+U\right)=\left(v+w\right)+U\right)\)

\(\lambda\cdot[v]=\left\lbrack\lambda\cdot v\right\rbrack\) \(\left(\lambda\left(v+U\right)=\lambda v+U\right)\)​

Check well definition(Take two equal element, then proof their \(+\) and \(\cdot\) are also equal)

If \([v]=[v^{\prime}]\) and \([w]=[w^{\prime}]\Rightarrow\left\lbrack v+w\right\rbrack=\left\lbrack v^{\prime}+w^{\prime}\right\rbrack\)

(\(v+w-(v^{\prime}+w^{\prime}))=v-v^{\prime}+w-w^{\prime}\Rightarrow\left\lbrack v+w\right\rbrack=\left\lbrack v^{\prime}+w^{\prime}\right\rbrack\)​

If \([v]=[v^{\prime}]\Rightarrow\left\lbrack\lambda v\right\rbrack=[\lambda v^{\prime}]\)

\((\lambda v-\lambda v^{\prime})=\lambda\left(v-v^{\prime}\right)\in U\Rightarrow\left\lbrack\lambda v\right\rbrack=\left\lbrack\lambda v^{\prime}\right\rbrack)\)

To check it just check their subtraction is in the \(U\)

Formal definition

Given a vector space \(V\) and a subspace \(U\), the quotient \(V/U\) is the vector space with underlying set \(\left\lbrace\left\lbrack v\right\rbrack:v\in V\right\rbrace\) and \(+,\cdot\) as defined in here

Check the vector space axioms: The \(0\) of \(V/U\) is \([0]=0+U=U\), also \(-[v]=[-v]\)

Proposition

Let \(U\hookrightarrow V\)

1. The quotient projection \(\pi:V\rightarrow V/U,v\mapsto[v]\) is linear

2. \(\dim (V/U)=\dim V-\dim U\)​

Proof

1. \(\pi(v+\lambda w)=[v+\lambda w]=[v]+\lambda[w]=\pi(v)+\lambda\pi(w)\)​

2. Dimension theorem \(\dim V=\dim(\text{Range}\pi)+\dim(\text{Null}\pi)=\dim V/U+\dim U\Rightarrow\dim V/U=\dim V-\dim U\)​

Theorem(The First Isomorphism Theorem)

Let \(T\in L(V,W)\). The map \(\bar T:V/\text{Null}T\rightarrow W\) defined by \(\bar{T}\left(v+\text{Null}T\right)=T\left(v\right)\) as a well defined linear map

\(\bar T\) is injective and \(\text{Range}(\bar{T})=\text{Range}(T)\) and \(\text{Range}T\simeq V/\text{Null}T\)

Proof

Well definition: \(v+\text{Null}T=v^{\prime}+\text{Null}T\Rightarrow v-v^{\prime}\in\text{Null}T\Rightarrow T\left(v-v^{\prime}\right)=0\Rightarrow T\left(v\right)=T\left(v^{\prime}\right)\Rightarrow\bar{T}\left\lbrack v+\text{Null}T\right\rbrack=\bar{T}\left\lbrack v^{\prime}+\text{Null}T\right\rbrack\)​

Injectivity: \(\bar{T}\left\lbrack v+\text{Null}T\right\rbrack=0\Rightarrow0=\bar{T}\left\lbrack v+\text{Null}T\right\rbrack=T\left(v\right)\Rightarrow v\in\text{Null}T\Rightarrow v+\text{Null}T=\text{Null}T\). Hence \(\bar T\) is injective

\(Range\bar{T}=\left\lbrace w=\bar{T}\left(v+\text{Null}T)=T\left(v\right)\right.\right\rbrace=Range\left(T\right)\)

By dim theorem, \(\dim(V/\text{Null}T)=\dim\text{Range}\bar{T}+\dim\text{Null}\bar{T}=\dim\text{Range}\bar{T}=\dim\text{Range}T\) since \(\bar T\) is injective, then the null space is \(\{0\}\)

Also \(\dim(V/\text{Null}T)=\dim V-\dim \text{Null}T\)

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