11.26 Isomorphisms

Representing linear maps by matrices

\(V\) with basis \(\mathcal{B}: v_1, \dots, v_n\), \(W\) with basis \(\mathcal{B}': w_1, \dots, w_m\).

For each \(T: V \to W\) linear map \(\mapsto [T]_{\mathcal{B} \mathcal{B}'} \in M_{m \times n}(\mathbb{F})\).

Recall that we have (coordinate):

\([~~]_{\mathcal{B}}:V \to\mathbb{F}^{n},\quad v\mapsto\left\lbrack v\right\rbrack_{\mathcal{B}}\text{ ,if }v=a_1v_1+\dots+a_{n}v_{n}\implies[v]_{\mathcal{B}}=(a_1,\dots,a_{n}),\\ \psi_{\mathcal{B}}:\mathbb{F}^{n} \to V,\quad(a_1,\dots,a_{n})\mapsto v=a_1v_1+\dots+a_{n}v_{n}\)​

Remark

They are inverses of each other.

Proposition

\([T \cdot v]_{\mathcal{B}'} = [T]_{\mathcal{B}\mathcal{B}'} [v]_{\mathcal{B}}\)

\(([~~]_{\mathcal{B}^{\prime}}\circ T)(v)=(T_{A}\circ[~~]_{\mathcal{B}})(v)\)

Diagram: ​

\("/// "\) means the diagram commutes.

Application

\(A=[T]_{\mathcal{BB}^{\prime}}\)

\(v \in \text{Null } T \iff A \cdot [v]_{\mathcal{B}} = 0\)

\(w \in \text{Range } T \iff A \cdot X = [w]_{\mathcal{B}'}\), for some \(X \in \mathbb{F}^m\)

Remark

1. Given \(T\), we get \(T_A: \mathbb{F}^n \to \mathbb{F}^m\), where \(A = [T]_{\mathcal{B}\mathcal{B}'}\).

2. Conversely: Given \(T_A\) (any \(A\)): \(\mathbb{F}^n \to \mathbb{F}^m\), we get a \(T: V \to W\) by \(T(v)=\psi_{\mathcal{B}^{\prime}}\left(T_{A}\cdot[v]_{\mathcal{B}}\right)\)​\([T = \psi_{\mathcal{B}'} \circ T_A \circ [~~]_{\mathcal{B}}]\)

Recall

\(\mathcal{L}(V, W) \to M_{m \times n}(\mathbb{F}) \to \mathcal{L}(V, W)\)

\(M_{m \times n}(\mathbb{F}) \to \mathcal{L}(V, W) \to M_{m \times n}(\mathbb{F})\) Both are the identities.

Proposition

Let \(V \xrightarrow{T} W \xrightarrow{S} Z(V \xrightarrow{ST} Z)\), and \(\mathcal{B}, \mathcal{B}'\), and \(\mathcal{B}''\) be bases of \(V, W\), and \(Z\), respectively.

Then \([ST]_{\mathcal{B}\mathcal{B}''} = [S]_{\mathcal{B}'\mathcal{B}''} \cdot [T]_{\mathcal{B}\mathcal{B}'}\) (\([ST]_{\mathcal{BB}''}\) is \(C\), \([S]_{\mathcal{B'B}''}\) is \(A\), \([T]_{\mathcal{BB}'}\) is \(B\),)

Proof.

Assume \(\mathcal{B}: v_1, \dots, v_n\), \(\mathcal{B}'': z_1, \dots, z_k\), \(\mathcal{B}': w_1, \dots, w_m\).

Consider \(([ST]_{\mathcal{BB}''})_{ij}:\)​ \(S(T(v_{j}))=S\left(\sum_{l=1}^{m}b_{lj}w_{l}\right)=\sum_{l=1}^{m}b_{lj}S(w_{l})=\sum_{l=1}^{m}b_{lj}\cdot\sum_{r=1}^{k}a_{rl}\cdot z_{r}=\sum_{r=1}^{k}\left(\sum_{l=1}^{m}b_{lj}a_{rl}\right)z_{r}\)

Hence, \(([ST]_{\mathcal{BB}''})_{ij}=\sum_{l=1}^{m}b_{lj}a_{il}=\sum_{l=1}^{m}a_{il}b_{lj}=(AB)_{ij}\)​\(\implies C_{ij} = (AB)_{ij} \implies C = AB.\)

Isomorphisms

Definition

A linear map \(T \in \mathcal{L}(V, W)\) is invertible if there is an \(S \in \mathcal{L}(W, V)\) such that \(ST = \text{Id}_V\) and \(TS = \text{Id}_W\). \(S\) is an inverse of \(T\).

Remark

1. If \(T\) is invertible, there is only one inverse, called \(T^{-1}\).

2. If \(T\) is invertible, \(T\) is injective and surjective.

Proposition

\(T \in \mathcal{L}(V, W)\) is invertible if and only if \(T\) is injective and surjective.

Proof

\(\Rightarrow\)) Since \(T\) is invertible, then there exists \(T^{-1}\) such that \(T^{-1}T = \text{Id}_V\) and \(TT^{-1} = \text{Id}_W\)

1. Injective

   Consider \(T(v_1)=T(v_2)\), then \(T^{-1}T(v_1)=T^{-1}T(v_2)\Rightarrow v_1=v_2\)​

2. Surjective

   We need to prove \(\forall w\in W,\exists v\in V:T(v)=w\), thus consider \(v=T^{-1}(w)\), then it is valid...

\(\Leftarrow\)) We need to prove \(\exists S\) such that \(ST = \text{Id}_V\) and \(TS = \text{Id}_W\).

Since \(T\) is bijective, then for any \(w\in W\) there exists a unique \(v\in V\) such that \(T(v)=w\)

Thus we define \(S(w)=v\left(v\text{ is unique}\right)\)

1. \(TS = \text{Id}_W\)

   \(TS(w)=T(S(w))=T(v)=w\)

2. \(ST = \text{Id}_V\)

   \(ST(v)=S(T(v))=S(w)=v\)​

3. Linear

   easy

Corollary

\(T \in \text{End}(V)\) is invertible if and only if \(T\) is injective if and only if \(T\) is surjective.

Definition

An invertible map \(T: V \to W\) is an isomorphism from \(V\) to \(W\). In this case, \(V\) and \(W\) are isomorphic. We write \(V\simeq W\).

Examples

1. \(\mathbb{C}_3\left\lbrack x\right\rbrack\simeq\mathbb{C}^4\)

   Define \(T:\mathbb{C}^4\to\mathbb{C}_3\left\lbrack x\right\rbrack\) by \((a, b, c, d) \mapsto a + bx + cx^2 + dx^3\).

   \(T\) is linear.

   Moreover: \(\text{Null } T = \{0\} \implies T\) is injective. \(\text{Range }T=\mathbb{C}_3\left\lbrack x\right\rbrack\implies T\) is surjective.

2. \(M_{2\times3}(\mathbb{Q})\simeq\mathbb{Q}^6\)

   \(T: M_{2 \times 3}(\mathbb{Q}) \to \mathbb{Q}^6\) by \(\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} \mapsto (a, b, c, d, e, f)\).

   \(T\) is linear and both injective and surjective.

   \(S: M_{2 \times 3}(\mathbb{Q}) \to \mathbb{Q}^6\) by \(\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} \mapsto (a, d, b, e, c, f)\).

   \(S\) is linear and both injective and surjective.

Example

\(\mathcal{L}(V, W) \simeq M_{m \times n}(\mathbb{F}) \quad (\dim V = n, \dim W = m)\). since

Notice that, for fixed bases \(\mathcal{B}\) and \(\mathcal{B}'\) of \(V\) and \(W\), respectively. We have a particular isomorphism. In general, all are different.

Remark

Being isomorphic is an equivalence relation.

1. \(V \simeq V\) (Id).

2. If \(V\simeq W\implies W\simeq V\) (\(T^{-1}\)).

3. If \(V\simeq W\) and \(W\simeq U\implies V\simeq U\) (\(ST\)).

Theorem (Classification Theorem)

Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

Proof

\(\Rightarrow\)) Let \(T:V\rightarrow W\) an isomorphism, \(T\) is injective and surjective, then by theorem

\(\dim V = \dim(\text{Null } T) + \dim(\text{Range } T) = 0 + \dim W\).

\(\Leftarrow\)) Let \(v_1, \dots, v_n\) be a basis of \(V\) and let \(w_1, \dots, w_n\) be a basis of \(W\). Let \(T\) be defined (Fundamental Theorem of Linear Maps) as \(T(v_i) = w_i\).

Claim: \(T\) is an isomorphism. To prove inj and sur, we need this.

1. \(\text{Range } T \supseteq \{w_1, \dots, w_n\} \implies \text{Range } T \supseteq \langle w_1, \dots, w_n \rangle = W\) (since \(w_1, \dots, w_n\) is a basis).

   Then \(\text{Range}T=W\) since \(W\subseteq \text{Range}T\) initially.

2. \(v \in \text{Null } T \implies 0 = T(a_1 v_1 + \dots + a_n v_n) = a_1 T(v_1) + \dots + a_n T(v_n)\) \(= a_1 w_1 + \dots + a_n w_n \implies a_1 = \dots = a_n = 0 \implies v = 0\).

\(\therefore T\) is injective and surjective. Thus, \(T\) is an isomorphism.

Remark

Every \(n\)-dimensional vector space \(V\) over \(\mathbb{F}\) is isomorphic to \(\mathbb{F}^n\).

1. Since \(\mathbb{F}^n \not\simeq \mathbb{F}^m\) if \(n \neq m\), so that every finite-dimensional (non-zero) \(\mathbb{F}\)-vector space is isomorphic to a unique \(\mathbb{F}^n\), namely \(\mathbb{F}^{\dim V}\).

2. \(\dim \mathcal{L}(V, W) = \dim M_{m \times n}(\mathbb{F}) = \dim \mathbb{F}^{m \cdot n} = m \cdot n\).

   Given bases \(v_1, \dots, v_n\) and \(w_1, \dots, w_m\) of \(V\) and \(W\), respectively, the list \(T_{ij}\) for \(i = 1, \dots, n\), \(j = 1, \dots, m\), defined by \(T_{ij}:\begin{cases}v_{i}\mapsto w_{j}\\ v_{l}\mapsto0,l\neq i{}\end{cases}\) is a basis of \(\mathcal{L}(V, W)\).

   For example \(\dim\mathcal{L}(R^2,R^2)=\dim M_{2\times2}\left(R\right)=\dim R^4=4\).

   The basis of \(R^2\) is \((1,0),(0,1)\) and \((1,1),(-1,1)\)​

   Then the basis of \(\mathcal{L}(R^2,R^2)\) is \(\left\lbrace T_{11}:\left(1,0\right)\mapsto\left(1,1\right),T_{12}:\left(1,0\right)\mapsto\left(-1,1\right),T_{21}:\left(0,1\right)\mapsto\left(1,1\right),T_{22}:\left(0,1\right)\mapsto\left(-1,1\right)\right\rbrace\)

   \(T(x,y)=(x−y,x+y)\) can be expressed as?

   \(T(x,y)=aT_{11}+bT_{12}+cT_{21}+dT_{22}\)​

   \(T(v_1) = T(1, 0) = (1 - 0, 1 + 0) = (1, 1)\)​\(T(v_2) = T(0, 1) = (0 - 1, 0 + 1) = (-1, 1)\)

   For \(T(v_1)\):

   \(T(v_1) = aT_{11}(v_1) + bT_{12}(v_1) + cT_{21}(v_1) + dT_{22}(v_1)\)​\(= a\mathbf{w_1} + b\mathbf{w_2} + c \cdot 0 + d \cdot 0\)​\(= a(1, 1) + b(-1, 1) = (a - b, a + b)\)For \(T(v_2)\):

   \(T(v_2) = aT_{11}(v_2) + bT_{12}(v_2) + cT_{21}(v_2) + dT_{22}(v_2)\)​\(= a \cdot 0 + b \cdot 0 + c\mathbf{w_1} + d\mathbf{w_2}\)​\(= c(1, 1) + d(-1, 1) = (c - d, c + d)\)

   \(\begin{cases}a-b=1\\ a+b=1\\ c-d=-1\\ c+d=1\end{cases}\)​

   \(T = 1 \cdot T_{11} + 0 \cdot T_{12} + 0 \cdot T_{21} + 1 \cdot T_{22} = T_{11} + T_{22}\)

3. Isomorphisms carry basis to basis: If \(T: V \to W\) is an isomorphism and \(v_1, \dots, v_n\) is a basis of \(V\), then \(Tv_1, \dots, Tv_n\) is a basis of \(W\).

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