12.3 The dual space

Definition

Dual space

A linear functional on \(\mathbb{F}\)-vector space \(V\) is an \(f\in L(V,\mathbb{F})\). The dual space of \(V\) is \(V^{*}=L\left(V,\mathbb{F}\right)\) (from vector to field)

\(f\) is a linear map from \(V\) to \(F\)

Remark

\(\dim V^*=\dim V\)

Example

1. \(f:\R^2 \rightarrow \R,f(x,y)=x-2y\)​

2. \(tr:M_{n\times n}\left(\mathbb{C}\right)\rightarrow\mathbb{C},A\mapsto tr\left(A\right)\)​

3. \(\pi_{i}:\mathbb{F}^{n}\rightarrow\mathbb{F},\left(x_1,x_2,\ldots,x_{n}\right)\mapsto x_{i}\)​

4. \(ev:\mathbb{R}_{n}[x]\rightarrow\mathbb{R},p\left(x\right)\mapsto p\left(1\right)\)​

Dual basis

Given \(v_1,...,v_n\) a basis of \(V\), the dual basis of it is \(\varphi_1,\ldots,\varphi_{n}\), \(\varphi_{i}\in V^{*}\) defined by \(\varphi_{i}\left(v_{j}\right)=\delta_{ij}=\begin{cases}1,i=j\\ 0,i\neq j\end{cases}\)

That is \(\varphi_{i}\) is a part(basis) of linear functional (\(\varphi:V\rightarrow F\))

We don't know what exact the \(\varphi_i\) is, but we have the property to know how this basis act on the basis of \(V\)

And \(\varphi=\sum_{j=1}^{n}a_{j}\varphi_{i}\in V^{*}\)

Lemma

The dual basis \(\varphi_1,...,\varphi_n\) of \(v_1,...,v_n\) is a basis of \(V^*\)

Proof

Since the length of dual basis is \(n\) same with the basis of \(V^*\), then we only need to prove dual basis is linearly independent.

Consider \(\lambda_1\varphi_1+\cdots+\lambda_{n}\varphi_{n}=0\) and \(v=a_1v_1+\cdots+a_{n}v_{n}\)

Then we evaluate \(\left(\lambda_1\varphi_1+\cdots+\lambda_{n}\varphi_{n}\right)\left(v_{i}\right)=0=\lambda_{i}\), then we have \(\lambda_1=...=\lambda_n=0\)​

Hence \(\varphi_1,...,\varphi_n\) is linearly independent

Example

1. \(f:\mathbb{R}^3\rightarrow\mathbb{R},\left(x,y,z\right)\mapsto x+y+z\). If \(\varphi_1,\varphi_2,\varphi_3\) is the dual basis of \(e_1,e_2,e_3\), then how \(\varphi_1,\varphi_2,\varphi_3\) compose \(f\)?

   \(f=a\varphi_1+b\varphi_2+c\varphi_3\), note: we have the property to evaluate, then \(f\left(x,y,z\right)=a\varphi_1\left(x,y,z\right)+b\varphi_2\left(x,y,z\right)+c\varphi_3\left(x,y,z\right)\)

   \(=a\varphi_1\left(xe_1+ye_2+ze_3\right)+b\varphi_1\left(xe_1+ye_2+ze_3\right)+c\varphi_1\left(xe_1+ye_2+ze_3\right)=ax+by+cz\)​

   Also \(f(x,y,z)=x+y+z\) by defined, then \(f=\varphi_1+\varphi_2+\varphi_3\)

   Consider the following basis of \(\R^3\): \((1,1,1),(1,1,0),(1,0,0)\) and \(\psi_1,\psi_2,\psi_3\) its dual basis \(f=3\psi_1+2\psi_2+\psi_3\)​

   \(f(1,0,0)=\left(3\psi_1+2\psi_2+\psi_3\right)\left(1,0,0\right)=3\psi_1\left(1,0,0\right)+2\psi_2\left(1,0,0\right)+\psi_3\left(1,0,0\right)=0+0+1=1\)​

   \(f(0,1,0)=3\psi_1\left(0,1,0\right)+2\psi_2\left(0,1,0\right)+\psi_3\left(0,1,0\right)\) since \((0,1,0)=(1,1,0)-(1,0,0)\)​

   Then \(=3\cdot 0+2\cdot 1+(-1)=1\)​

Dual map

Given \(T\in L(V,W)\), the dual map of \(T\) is \(T^*\in L(W^*,V^*)\) defined by \(T^*(\varphi)=\varphi\circ T\) \([T^{*}(\varphi)\left(v\right)=\left(\varphi\circ T\right)\left(v\right)=\varphi\left(T\left(v\right)\right)]\)​

\(\varphi\in W^{*},\varphi:W\rightarrow\mathbb{F},\varphi\circ T:V\rightarrow W\rightarrow\mathbb{F},\varphi\circ T\in V^{*}\)​

Check: \(T^*\) is linear

\(T^{*}\left(\varphi+\lambda\psi\right)=\left(\varphi+\lambda\psi\right)\circ T=\varphi\circ T+\lambda\psi\circ T=T^{*}\left(\varphi\right)+\lambda T^{*}\left(\psi\right)\)

‍

Properties of the map \(-^*:T\rightarrow T^*\) \((L(V,W)\xrightarrow{-^*}L(W^{*},V^{*}))\)

\((T+S)^{*}=T^{*}+S^{*}\) and \((\lambda T)^*=\lambda T^*\), thus \(-^{*}\) is linear

Proof

\((T+S)^{*}\left(\varphi\right)=\varphi\circ\left(T+S\right)=\varphi\circ T+\varphi\circ S=T^*(\varphi)+S^*(\varphi)\), then \((T+S)^{*}=T^{*}+S^{*}\).why? ()

‍

\((\varphi_0 \circ (T+S))(v) = \varphi \left( [(T+S)(v)] \right) = \varphi \left( T(v) + S(v) \right)\)

\(= \varphi(T(v)) + \varphi(S(v))\)

\(= (\varphi_0 T)(v) + (\varphi_0 S)(v)\)

\(= (\varphi_0 T + \varphi_0 S)(v), \quad \forall v \in V\)

\(\implies \varphi_0 \circ (T+S) = \varphi_0 T + \varphi_0 S.\)

‍

\((\lambda T)^*(\varphi)=\varphi\circ (\lambda T)=\lambda\varphi(T)=\lambda T^*(\varphi)\)

Moreover claim \((TS)^{*}=S^*T^*\)

\((TS)^{*}\left(\varphi\right)=\varphi\circ\left(TS\right)=\left(\varphi\circ T\right)\circ S=S^{*}\left(\varphi\circ T\right)=S^{*}\left(T^{*}\left(\varphi\right)\right)\)​

Null\(T^*\) and Range\(T^*\)

Given a subspace \(U\) of \(V\), its annihilator is the subspace of \(V^*,U^\circ=\{\varphi\in V^*:\varphi(u)=0,\forall u\in U\}\)

Check: \(U^{\circ}\) is a subspace

\(V^{\circ}=\left\lbrace0\right\rbrace,\left\lbrace0\right\rbrace^{\circ}=V^*\)

Example

1. \(V=\mathbb{R}^3,U=\{(x,y,z):2x-y+z=0\},U^{\circ}=?\)​

   \(U=\langle\left(\frac12,1,0\right),\left(-\frac12,0,1\right)\rangle\) \(V=U\oplus\langle\left(0,0,1\right)\rangle\) we call this \(v\) and we call \(\left(\frac12,1,0\right),u_1\) and \(\left(-\frac12,0,1\right),u_2\)

   \(\varphi\in U^\circ\iff \varphi(u_1)=\varphi(u_2)=0\). Hence \(U^{\circ}=\left\lbrace\varphi\in V^{*}:\varphi(u_1)=\varphi(u_2)=0,\varphi\left(v\right)=t\right\rbrace\)

   Now, let's compute \(\varphi(1,0,0)=\varphi\left(-2\left(-\frac12,0,1\right)+2\left(0,0,1\right)\right)=2t\)​

   \(\varphi(0,1,0)=\varphi\left(\left(\frac12,1,0\right)+\left(-\frac12,0,1\right)-\left(0,0,1\right)\right)=-t\)

   \(\varphi(1,0,0)=t\)​

   \(\varphi(x,y,z)=\varphi\left(x\left(1,0,0\right)+y\left(0,1,0\right)+z\left(0,0,1\right)\right)=x2t-yt+zt=t\left(2x-y+z\right)\)

2. \(U=\left\lbrace A\in M_{2\times2}\left(\mathbb{C}\right):trA=0\wedge a_{21}=0\right\rbrace,U=\langle\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\rangle;V=U\oplus\langle\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\rangle\)​

   \(\varphi\in U^{\circ}\;\;\iff\;\;\varphi\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}=0=\varphi\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\)​

   \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\left(a+d\right)\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}+b\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+c\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+\left(-d\right)\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\;\)​

   Then \(\varphi\begin{pmatrix}a & b\\ c & d\end{pmatrix}\;\iff\;\;\;\left(a+d\right)s+ct\)​

Proposition

\(\dim U^\circ =\dim V-\dim U\)

Proof 1 Following the strategy in the examples above

Since \(U\) is a subspace of \(V\), then we have \(V=U\oplus W\).

Suppose the basis of \(U\) is \(\{u_1,...,u_n\}\) and the basis of \(W\) is \(\{w_1,...,w_k\}\)

Then the basis of \(V\) is \(\{u_1,...,u_n,w_1,...,w_k\}\). Then let's take any \(v\in V\)

Since \(\varphi\in U^\circ\), then consider \(\varphi(v)=a_1\varphi(u_1)+\cdots+a_{n}\varphi(u_{n})+a_{n+1}\varphi(w_1)+\cdots+a_{n+k}\varphi(w_{k})=a_{n+1}\varphi(w_1)+\cdots+a_{n+k}\varphi(w_{k})\)

Thus \(\dim U^\circ=\dim W\). Since we know \(\dim V=\dim U +\dim W\), then \(\dim U^\circ =\dim V-\dim U\)

Proof 2

Let \(inc:U\rightarrow V\) and consider \(inc^*:V^*\rightarrow U^*\). We have \(\dim V^{*}=\dim\left(\text{Null}inc^{*}\right)+\dim(\text{Range}inc^*)\)​

\(\text{Null}inc^{*}=\{\varphi\in V^*:i^*(\varphi)=\varphi\circ inc=0\}=U^\circ\) and \(\text{Range}inc^{*}=\{\psi\in V^{*}:\psi=inc^*(\psi)=\psi\circ inc\}=U^*\)​

Hence \(\dim V=\dim V^{*}=\dim\left(\text{Null}inc^{*}\right)+\dim(\text{Range}inc^{*})=\dim U^{\circ}+\dim U^{*}=\dim U^{\circ}+\dim U\)​

Theorem

Let \(V\) and \(W\) be finite-dimensional vector spaces and let \(T\in L(V,W)\). Then:

1. \(\text{Null}T^*=(\text{Range}T)^\circ\)​

2. \(\dim \text{Null}T^*=\dim \text{Null}T+\dim W-\dim V\)​

3. \(\dim \text{Range}T^*=\dim(\text{Range}T)\)​

4. \(\text{Range}T^{*}=(\text{Null}T)^{\circ}\)

Proof

1. \(\subseteq\)) \(0=T^{*}(\varphi)=\varphi\circ T\Rightarrow\left(\varphi\circ T\right)\left(v\right)=0,\forall v\in V\Rightarrow\varphi\left(\text{Range}T\right)=0\Rightarrow\varphi\in\left(\text{Range}T\right)^{\circ}\)​

   \(\supseteq\)) \(\varphi\in\left(\text{Range}T\right)^{\circ}\Rightarrow\varphi\left(T\left(v\right)\right)=0,\forall v\in V\Rightarrow\varphi\circ T=0\Rightarrow T^*(\varphi)=0\Rightarrow \varphi\in \text{Null}T^*\)​

2. \(\dim\text{Null}T^{*}=\dim(\text{Range}T)^{\circ}=\dim W-\dim(\text{Range}T)=\dim W-\left(\dim V-\dim\text{Null}T\right)=\dim\text{Null}T+\dim W-\dim V\)​

3. \(\dim(\text{Range}T^{*})=\dim W^{*}-\dim\text{Null}T^{*}=\dim W^{*}-(\dim W-\dim\text{Range}T)=\dim\text{Range}T\)​

4. Notice that \(\dim(\text{Null}T)^\circ=\dim V-\dim \text{Null}T=\dim (\text{Range}T)=\dim(\text{Range}T^*)\)​

   So it's suffices to prove that \(\text{Range}T^{*}\subseteq(\text{Null}T)^{\circ}\)​

   \(\psi\in\text{Range }T^{*},\quad\psi=T^{*}(\varphi)=\varphi_0T.\)​

   Hence, \(\psi(v) = (\varphi_0 T)(v) = \varphi(T(v)).\)

   So, if \(T\left(v\right)\left.\implies\psi(v\right)=0\implies\psi\in\left(\text{Null}T\right)^{\circ}.\)

Corollary

It holds:

(a) \(T\) is surjective iff \(T^*\)​ is injective.

(b) \(T\) is injective iff \(T^*\) is surjective.

Proposition

\([T^*]=[T]^t\)

m>n, T 是长方形 T^* 就要transpose

actually we need to choose basis to write the T matirx and use dual basis o write T^*