# Random math: 17

Source of most content: Gunhee Cho

# Dual space

Def. Given a finite dimensional vector space $V$ over $F$, Define $V^* := \mathcal{L}(V, F) = \{ f: V \rightarrow F \mid f \text{ is linear} \}$, which is called the dual space of $V$.

NB1; $V^*$ is a VS over $F$ when we equipped with $\forall f,g \in V^*,\ (f+g)(v) := f(v) + g(v)), \forall c \in F, (cf)(v) := cf(v)$.

Given finite dimensional vector spaces $V,W$ over $F$, consider $\mathcal{L}(V, W) = \{ f: V \rightarrow W \mid \text{ is linear}\}$. Let $dim V = n, \ dim W = m$. Choose a basis $\beta$ for $V$ and $\gamma$ for $W$. Define $\Phi: \mathcal{L}(V,W) \rightarrow Mat_{n \times m}(F) \ (T \mapsto [ T ] _\beta^\gamma)$. Then, $\mathcal{L}(V, W) \stackrel{\Phi}{\cong} Mat _{n \times n}(F).$ Therefore, $dim \mathcal{L}(V, W) = nm$. In particular, if $W = F$, then, $V ^*=\mathcal{L}(V,F)$ has same dimension with $V$, i.e., $dim V^ * = dim V$.

NB2; If $dim V = \infty,$ then $dim V^ * \ne dim V$ in general.

Def. Given a finite dimensional vector space $V$ over $F$, $V^* := \mathcal{L}(V, F) = \{ f: V \rightarrow F \mid f \text{ is linear} \}$, let $\beta =\{ v_1, \cdots, v_n \}$ be a basis for $V$. Define $v_i^*: V \rightarrow F$ s.t. $v_i^ *(v_j) = \delta_{ij},$ where $\delta_{ij}$ is Kronecker delta. Then, $\gamma = \{v_i^ * \mid 1 \leq i \leq n\}$ is a basis for $V^ *$, which is called the dual basis for $\beta$.

1. (linearly independent) $\sum_{i=1}^n a_i v_i^ * =0,$ i.e., $\forall v \in V, \sum_{i=1}^n a_i v_i^ *(v) =0$. Take $v = v_j,$ then $\sum_{i=1}^n a_i v_i^ *(v_j) = a_j = 0 \forall j.$ Therefore, $\gamma$ is linearly independent.
2. (span) For any $g \in V^*,$ let $b_i = g(v_i)$. Then, $\forall v = \sum_{i=1}^n c_i v_i \in V, \sum_{i=1}^n b_i v_i^ * (v) = \sum_{i=1}^n b_i v_i^ * (\sum_{j=1}^n c_j v_j)=\sum_{i=1}^n b_i c_i = \sum_{i=1}^n c_i g(v_i) = g(\sum_{i=1}^n c_i v_i) = g(v).$
Therefore, $span(\gamma) = V^ *$.

NB3; Let $f\in V^ *$, i.e., $f=\sum_{i=1}^n a_i v_i^ *$. Then, $f(v_k) = \sum_{i=1}^n a_i v_i^ * (v_k) = a_k$. Thus, $f= \sum_{i=1}^n f(v_i) v_i^ *$.
NB4; Dual basis depends on the basis $\beta$ of $V$.
NB5; Given a finite dimensional vector space, $V \rightarrow V^ * \rightarrow (V^ *)^ * \cdots,$ where all $V, V^ *, (V^ *)^ *$ is a vector space over $F$ and $dimV =dim V^ * = dim (V^ *)^ *$ (VS isomorphic).

## Double dual

Given finite dimensional vector spaces $V,W$ over $F$ with $dim V = dim W =n$, let $\beta = \{ v_1, \cdots, v_n \}$ be a basis of $V$ and $\gamma = \{ w_1, \cdots, w_n \}$ be a basis of $W$. Define $\Phi: V \rightarrow W \ (\sum_{i=1}^n a_i v_i \mapsto \sum_{i=1}^n a_i w_i).$

1. In case of $W=V^*$, $\gamma = \{ v_1^ *, \cdots, v_n^ * \}$, and $\Phi: V \mapsto V^ * (\sum_{i=1}^n a_i v_i \mapsto \sum_{i=1}^n a_i v_i^ *)$.
2. In case of $W= (V^ *)^ *$, define $\varphi: V \rightarrow (V^ *)^ * \ (v \mapsto \varphi(v))$ s.t. $\varphi(v): V^ * \rightarrow F \ (f \mapsto \varphi(v)(f):=f(v)),$ where $f:V \rightarrow F$.

Then, we can show that $\varphi$ is a vector space isomorphism.

1. (Linearity) $\forall v,w \in V,\forall c \in F,\forall f \in V^ * , \ \varphi(cv+w)(f) \stackrel{def}{=} f(cv+w) \stackrel{f:linear}{=} cf(v)+f(w) \stackrel{def}{=} c\varphi(v)(f)+\varphi(w)(f).$ Therefore, $\forall v,w \in V,\forall c \in F, \ \varphi(cv+w) = c\varphi(v) + \varphi(w)$.
2. (Bijective) By rank-nullity theorem, it is enough to show that $\varphi$ is injective since dimension is the same, i.e., show $\ker \varphi = \{ 0\}$. Let $\varphi(v) =0,$ i.e., $\forall f \in V^ *, \varphi(v)(f) = f(v) =0$. Let $\beta = \{ v_1, \cdots, v_n \}$ be a basis for $V$. Then, $\gamma = \{ v_1^ *, \cdots, v_n^ * \}$ be the dual basis for $\beta$. Then, $\forall f \in V^ *, \ f=\sum_{i=1}^n f(v_i)v_i^ * \leadsto f(v) = \sum_{i=1}^n f(v_i)v_i^ * ( \sum_{j=1}^n b_j v_j).$ Since this holds for all $f \in V^ *$, taking $f=v_i^ * \leadsto v_i^ *(\sum_{j=1}^n b_j v_j )=b_i =0, \forall i$. This implies that $v=0$, i.e., $\ker \varphi = \{ 0\}$.

NB6; Differently from dual space $V^ *$, double dual $V^{ * *}$ has natural isomorphism from $V$ onto $V^{ * * }$ since $\varphi$ does not depend on the choice of the basis of $V$.

# Dual of linear map

Def. Given finite dimensional vector spaces $V, W$ over $F$, let $T: V \rightarrow W$ be a linear operator. Define ${}^{t}T = W^ * \rightarrow V^ * \ (g \mapsto {}^{t}T (g) = g \circ T \in V^ *),$ which is called the dual of $T$ (transpose of $T$).

NB7; Since we define the addition and scalr multiplication in NB1, $\forall c \in F, \forall g,h \in W^ *, \ {}^{t}T(cg+h) \stackrel{def}{=} (cg+h)(T) = cg(T) + h(T) = c{}^{t}T(g) +{}^{t}T(h)$, i.e., ${}^{t}T$ is linear.

Def. Given a finite dimensional inner product space $(V, \langle , \rangle)$, let an linear operator $T:V\rightarrow T$. Define $T^ *: V \rightarrow V$ s.t. $\forall x,y \in V, \ \langle T(x),y \rangle = \langle x, T^ *(y) \rangle,$ which is called the adjoint operator of T.

NB8; $\exists! T^ *$ when we consider a finite dimensional space.

## Proposition

$\text{Let } \beta = \{ v_1, \cdots, v_n \} \text{ be a basis of } V, \gamma=\{ w_1, \cdots, w_n \} \text{ for } W. \text{ Choose dual bases } \beta ^*, \gamma^ * \text{ for } V^ *, W^ *. \nonumber$ $\text{ Then, } [ {}^{t}T ] _{\gamma^ *}^{\beta^ *} = ([ T ] _{\beta}^{\gamma})^\top. \nonumber$

This is why ${}^{t}T$ is called transpose of $T$.

### Proof

Let $[ T ] _{\beta}^{\gamma} = (a _{ij}) \in Mat _{m\times n}(F)$ and $[ {}^{t}T ] _{\gamma^ *}^{\beta^ *} = (b _{ij}) \in Mat _{n\times m}(F)$. Then, ${}^{t}T (w_j^ *)= \sum _{k=1}^n b _{kj} v _k^ *.$ At first, ${}^{t}T (w _j^ *)(v _i) = \sum _{k=1}^n b _{kj}v_k^ * (v _i) = b _{ij}$. Also, since $T(v _i)=\sum _{k=1}^m a _{ki} w _k$, ${}^{t}T (w _j^ *)(v _i) = (w _j^ * \circ T)(v _i) = w _j^ * (T(v _i))= w _j^ * (\sum _{k=1}^m a _{ki} w _k)=\sum _{k=1}^m a _{ki} w _j^ * (w _k) = a _{ji}.$ Thus, $b _{ij} = a _{ji}$.

$\text{For an linear operator } T: V\rightarrow V, \ \exists! T^ *: V \rightarrow V \text{ s.t. }\forall x,y \in V, \ \langle T(x),y \rangle = \langle x, T^ *(y) \rangle. \nonumber$

### Proof

1. (Existence) For $y \in V$, define $g:V \rightarrow F \ (x \mapsto \langle T(x),y \rangle)$. Then, $g$ is linear since both $T$ and inner product is linear (w.r.t 1st component). By Riesz representation, $\exists! v_y \in V,$ s.t. $g(\cdot) = \langle \cdot ,v_y \rangle.$ Define $T^ * : V \rightarrow V \ (y \mapsto v_y)$. Then, we need to show that $T^ *$ is (a) linear and (b) satisfy $\forall x,y \in V, \ \langle T(x),y \rangle = \langle x, T^ *(y) \rangle$.
• (a) $T^ *(y_1 + cy_2) = v_{y_1 + cy_2} \stackrel{NTS}{=} T^ *(y_1) + cT^ *(y_2)= v_{y_1} + cv_{y_2}.$ So we need to show $\langle \cdot, v_{y_1 + cy_2} \rangle = \langle \cdot, v_{y_1} + cv_{y_2} \rangle$. $\forall x \in V, \langle x, v_{y_1 + cy_2} \rangle \stackrel{def}{=} \langle T(x), y_1 + cy_2 \rangle = \langle T(x), y_1\rangle + \bar{c}\langle T(x), y_2 \rangle$ $= \langle x, v_{y_1} \rangle + \bar{c}\langle x, v_{y_2} \rangle = \langle x, v_{y_1} \rangle + \langle x, cv_{y_2} \rangle$. Therefore, $T^ *$ is linear.
• (b) $g(x) = \langle x, v_{y} \rangle = \langle x, T^ *(y) \rangle = \langle T(x), y \rangle$. Since $y$ is arbitrary, we have $\langle x, T^ *(y) \rangle = \langle T(x), y \rangle, \ \forall x,y \in V$.
2. (Uniqueness) If $\tilde{T}: V \rightarrow T$ satisfies $\langle x, \tilde{T}(y) \rangle = \langle T(x), y \rangle, \ \forall x,y \in V$. Then, $\langle x, T^ *(y)-\tilde{T}(y) \rangle$ holds for all $x \in V$. Then, take $x = T^ *(y)-\tilde{T}(y) \leadsto \Vert T^ *(y)-\tilde{T}(y) \Vert =0,$ i.e., $T^ *(y)=\tilde{T}(y), \forall y$.

$\text{Given } (V, \langle , \rangle), \text{ Let } \beta \text{ be orthonormal basis for }V. \text{ Let } T:V\rightarrow V. \text{ Then, } [ T^ *]_\beta = ([ T] _\beta)^H. \nonumber$

### Proof

Let $\beta=\{ v_1, \cdots, v_n\}$ be orthonormal basis. Then, by Gram–Schmidt, $T^ *(v_j) = \sum_{i=1}^n \langle T^ *(v_j), v_i \rangle v_i \stackrel{adjoint}{=} \sum_{i=1}^n \langle v_j, T(v_i) \rangle v_i = \sum_{i=1}^n \overline{\langle T(v_i), v_j \rangle} v_i$, i.e., $\overline{\langle T(v_i), v_j \rangle}$ is $(i,j)$ element $(b_{ij})$ of $[ T^ *] _\beta$. On the other hand, $T(v _i) = \sum _{j=1}^n \langle T(v _i), v _j \rangle v _j,$ i.e., $\langle T(v _i), v _j \rangle$ is $(j,i)$ element $(a _{ji})$ of $[ T] _\beta$. Thus, $b _{ij} = \overline{\langle T(v_i), v_j \rangle} = \overline{a _{ji}}.$

$T=L_A: F^n \rightarrow F^n \ (X \mapsto AX). \text{ Then, } (L_A)^ * = L_{A^H}. \nonumber$

### Proof

Let $\epsilon_n = \{ e_1, \cdots, e_n \}$ be the standard orthonormal basis for $F^n$. Then, $[ (L _A)^ * ] _{\epsilon _n} = ([ L _A ] _{\epsilon _n})^H$ by above corollary. Since $A= [ L _A ] _{\epsilon _n}$, $A^H \stackrel{def}{=} [ L _{A^H} ] _{\epsilon _n} = [ (L _A)^ * ] _{\epsilon _n}$. Therefore, $( L _A)^ * = L _{A^H}$.

$\text{For linear operator }T: V\rightarrow V, \text{ Following holds: } \nonumber$
1. $T_1, T_2:V\rightarrow V,\ (T _1 + T _2 )^ * = T _1^ * + T _2^ *$.
2. For $c \in F$, $(cT)^ * = \bar{c}T^ *$.
3. $(T_1 \circ T_2)^ * = T_2^ * \circ T_1^ *$.
4. $(T^ *)^ * = T$.
5. $Id^ * = Id$.

### Proof

1. $\forall x,y \in V, \ \langle (T_1 + T_2)(x), y \rangle \stackrel{def}{=} \langle x, (T_1^ * + T_2^ *)(y) \rangle.$ Here, $\langle (T_1 + T_2)(x), y \rangle = \langle T_1(x) + T_2(x), y \rangle$ $= \langle T_1(x), y \rangle + \langle T_2(x), y \rangle \stackrel{adjoint}{=} \langle x, T _1^ * (y) \rangle + \langle x, T_2^ * (y) \rangle = \langle x, T _1^ * (y) + T_2^ * (y) \rangle$. Therefore, $(T_1^ * + T_2^ *)(y) = T _1^ * (y) + T_2^ * (y), \forall y$, i.e., $(T _1 + T _2 )^ * = T _1^ * + T _2^ *$.
2. $\forall x,y \in V, \ \langle (cT)(x), y \rangle = \langle x, (cT)^ * (y) \rangle$. Here, $\langle (cT)(x), y \rangle = \langle cT(x), y \rangle = c \langle x, T^ *(y) \rangle = \langle x, \bar{c} T^ *(y) \rangle$. Therefore, $(cT)^ * = \bar{c}T^ *$.
3. $\forall x,y \in V, \ \langle (T_1 \circ T_2)(x), y \rangle = \langle x, (T_1 \circ T_2)^ *(y) \rangle$. Here, $\langle (T_1 \circ T_2)(x), y \rangle = \langle (T_1)(T_2(x)), y \rangle \stackrel{T_2(x) \in V}{=} \langle T_2(x), T_1^ * (y) \rangle \stackrel{T_1(x) \in V}{=} \langle x, T_2^ * ( T_1^ * (y)) \rangle = \langle x, T_2^ * \circ T_1^ * (y) \rangle.$
4. $\forall x,y \in V, \ \langle (T^*)(x), y \rangle = \langle x, (T^ *)^ *(y) \rangle$. Here, $\langle (T^ *)(x), y \rangle = \overline{\langle y, (T^ *)(x) \rangle} \stackrel{adjoint}{=} \overline{\langle T(y), x \rangle} = \langle x, T(y) \rangle.$
5. $\forall x,y \in V, \ \langle Id(x), y \rangle = \langle x, Id^ * (y) \rangle$. Here, $\langle Id(x), y \rangle = \langle x, y \rangle$.