12.6 Dual space

1. Let \(\mathcal{B} = \{(1, 0, 1), (0, 1, -2), (-1, -1, 0)\} = \{v_1, v_2, v_3\}\) be a basis for \(\mathbb{R}^3\).

   1. Describe the dual basis \(\mathcal{B}^*\) of \(\mathcal{B}\) (explicitly).

      \(\mathcal{B}^* = \{\varphi_1, \varphi_2, \varphi_3\}\) where \(\varphi_{i}(v_{j})=\begin{cases}1 & \text{if }i=j\\ 0 & \text{if }i\neq j\end{cases}\)

      So if we have \((x, y, z) = a v_1 + b v_2 + c v_3\)​\(\implies \varphi_1(x, y, z) = a\), \(\varphi_2(x, y, z) = b\), and \(\varphi_3(x, y, z) = c\).

      We want to write \((x, y, z) = a v_1 + b v_2 + c v_3\).

      \(\begin{aligned}\begin{bmatrix}1 & 0 & -1\\ 0 & 1 & -1\\ 1 & -2 & 0\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}x\\ y\\ z\end{bmatrix}\implies\text{Row-reduce}\implies\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}-z+2x-2y\\ -z+x-y\\ -z+x-2y\end{bmatrix}.\end{aligned}\)

      Therefore:
      ​\(\varphi_1(x, y, z) = a = -z + 2x - 2y\) \(\varphi_2(x, y, z) = b = -z + x - y\) \(\varphi_3(x, y, z) = c = -z + x - 2y\). 2. Let \(f \in (\mathbb{R}^3)^*\) such that \(f(v_1) = 1\), \(f(v_2) = -1\), \(f(v_3) = 3\). Find \(f(x, y, z)\) and give its coordinates in \(\mathcal{B}^*\).

      We have that \(f(x, y, z) = f(a v_1 + b v_2 + c v_3) = a f(v_1) + b f(v_2) + c f(v_3)\).

      \(\begin{aligned}f(x,y,z)=a-b+3c=4x-7y-3z\end{aligned}\)

      Then we have that \(f(x,y,z)=a-b+3c=\varphi_1(x,y,z)-\varphi_2(x,y,z)+3\varphi_3(x,y,z)\)​

      So the coordinates of \(f\) in \(\mathcal{B}^* = \{\varphi_1, \varphi_2, \varphi_3\}\) are: \(\begin{bmatrix}f\end{bmatrix}_{\mathcal{B}^{*}}=\begin{bmatrix}1\\ -1\\ 3\end{bmatrix}.\)

2. Every linear functional \(\varphi: V \to \mathbb{F}\) is either the \(0\) map or surjective.

   Note that \(\dim(\mathbb{F}) = 1\) (\(\{1\}\) is a basis for \(\mathbb{F}\)). So if we have a subspace \(S\) of \(\mathbb{F}\):
   \(\dim(S) \leq \dim(\mathbb{F}) = 1 \implies \dim(S) = 0 \text{ or } \dim(S) = 1\)​\(\implies S = \{0\} \text{ or } S = \mathbb{F}\).

   Now, since \(\text{Range}(\varphi)\) is a subspace of \(\mathbb{F}\): \(\text{Range}(\varphi) = \{0\} \text{ or } \text{Range}(\varphi) = \mathbb{F}\) \(\implies \varphi = 0 \text{ or } \varphi \text{ is surjective.}\)

3. Suppose \(W\) is finite-dimensional and \(T \in L(V, W)\). Prove that \(T^* = 0 \iff T = 0\).

   \(\implies\) If \(T^* = 0\), let \(\varphi \in W^*\), \(T^*(\varphi) = 0\). So if \(v \in V\), \(0 = T^*(\varphi)(v) = \varphi(T(v))\).

   Thus, \(0 = \varphi(T(v))\) \(\forall \varphi \in W^*\). Since \(\varphi\) is arbitrary, \(T(v) = 0\).

   So \(T(v) = 0\) \(\forall v \in V \implies T = 0\)

   \(\impliedby\) If \(T = 0\), take \(\varphi \in W^*\), \(T^*(\varphi) = \varphi \circ T = \varphi \circ 0 = 0\). (Since \(\varphi(T(v)) = \varphi(0) = 0\).) Therefore \(T^* = 0\)

4. Define \(T: \mathbb{R}^3 \to \mathbb{R}^2\), \(T(x, y, z) = (4x + 5y + 6z, 7x + 8y + 9z)\). Suppose \(\{\psi_1, \psi_2\}\) and \(\{\varphi_1, \varphi_2, \varphi_3\}\) are the dual bases of the canonical bases of \(\mathbb{R}^2\) and \(\mathbb{R}^3\), respectively.

   1. Describe the linear functionals \(T^*(\psi_1)\) and \(T^*(\psi_2)\).

      The canonical basis for \(\mathbb{R}^2\) is \(\{(1, 0), (0, 1)\} = \{e_1, e_2\}\), and \(\psi_1(e_{i})=\begin{cases}1 & i=1\\ 0 & i=2\end{cases},\psi_2(e_{i})=\begin{cases}1 & i=2\\ 0 & i=1\end{cases}.\)

      So \(T^*(\psi_1)(x, y, z) = \psi_1(T(x, y, z)) = \psi_1(4x + 5y + 6z, 7x + 8y + 9z) = \psi_1(e_1(4x + 5y + 6z) + e_2(7x + 8y + 9z)) = 4x + 5y + 6z\).

      Similarly, \(T^*(\psi_2) = 7x + 8y + 9z\) (2nd coordinate of \(T(x, y, z)\)). 2. Write \(T^*(\psi_1)\) and \(T^*(\psi_2)\) as a linear combination of \(\{\varphi_1,\varphi_2,\varphi_3\}\) (dual basis of \(\mathbb{R}^3\)).

      Note that \(\varphi_1,(x,y,z)=\varphi_1(xe_1+ye_2+ze_3)=x,\quad\varphi_2(x,y,z)=y,\quad\varphi_3(x,y,z)=z.\)​

      So \(T^{*}(\psi_1)(x,y,z)=4x+5y+6z=4\varphi_1+5\varphi_2+6\varphi_3\)
      ​\(T^{*}(\psi_2)(x,y,z)=7x+8y+9z=7\varphi_1+8\varphi_2+9\varphi_3\)

5. Suppose \(V\) and \(W\) are finite-dimensional. Prove that the map that takes \(T \in \mathcal{L}(V, W)\) to \(T^* \in \mathcal{L}(W^*, V^*)\) is an isomorphism between \(\mathcal{L}(V, W)\) and \(\mathcal{L}(W^*, V^*)\).

   The map \(\Phi:\mathcal{L}(V,W)\to\mathcal{L}(W^{*},V^{*})\), \(\Phi\left(T\right)=T^{*}\) is an isomorphism.

   1. \(\Phi\) is a linear map:
      \(\Phi(T+\lambda S)=(T+\lambda S)^{*}=\left(lecture,exercise)=T^{*}\right.+\lambda S^{*}=\Phi(T)+\lambda\Phi(S)\)
   2. \(\Phi\) is injective: \(\text{Null}\Phi=\{0\}\;\iff\;\{T=0\iff T^*=0\}\) and this is proved in exercise 3
   3. \(\Phi\) is surjective: \(\dim(\mathcal{L}\left(V,W\right))=\dim(V)\cdot\dim(W)=\dim(V^{*})\cdot\dim(W^{*})=\dim(\mathcal{L}\left(W^{*},V^{*}\right))\).

   Thus, \(\Phi\) is an isomorphism.

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