# Notes - Linear Algebra MT23, Dual spaces

### Flashcards

Suppose $V$ is a vector space over $\mathbb F$. Can you define it’s dual $V’$?

Suppose $V$ is a vector space over $\mathbb F$. What is the name for functions $f : V \to \mathbb F$ where $f$ linear (i.e. elements of the dual space)?

Linear functionals.

Suppose $V$ is a finite dimensional vector space with a basis $\{e _ 1, \ldots, e _ n\}$. Can you define the dual basis?

where

\[e'_i : e_j \mapsto \begin{cases} 1 &\text{if } i =j \\\\ 0 &\text{otherwise} \end{cases}\]Let $T : V \to W$ be a linear map. Can you define the dual map $T’$, and why this definition makes sense?

where

\[T'f = f \circ T\]Consider triangular diagram with $V, W, \mathbb F$ at corners, with $T$ joining $V$ and $W$, $f \in W’$ joining $W$ and $\mathbb F$ and $f \circ T$ joining $V$ and $\mathbb F$.

Let $V$ and $W$ be two finite dimensional vector spaces. What natural isomorphism is there between $\text{Hom}(V, W)$ and $\text{Hom}(W’, V’)$?

Can you define $E _ v : V’ \to \mathbb F$, the evaluation map with respect to $v$, and how this relates to $V’’$?

This is an element of $V’’$.

Can you give a natural linear isomorphism (one that does not depend on a choice of basis) between $V$ and $V’’$?

where $E _ v : V’ \to \mathbb F \in V’’$

\[E_v(f) = f(v)\]Suppose $V$ is a vector space. What is a hyperplane?

The preimage $f^{-1}(\{c\})$ for a constant $c \in \mathbb F$ where $f \in V’$.

Let $T : V \to W$ be a linear map. The dual map $T’$ is defined as

\[T' : W' \to V'\]
where

\[T'f = f \circ T\]
Quickly prove that the matrix of $T’$ corresponds to the matrix of $T^\intercal$.

Let $A$ be the matrix of $T$ with respect to $[e _ 1, \cdots, e _ n]$, and let $B$ be the matrix of $T’$ with respect to the corresponding dual basis $[e _ 1’, \cdots, e _ n’]$. Then

\[T(e_ j) = \sum^{n}_ {i=1} a_ {ij} e_ i\]so

\[e'_ i(T(e_ j)) = a_ {ij}\]Also,

\[T'(e_ i') = \sum^n_ {j = 1} b_ {ji} e'_ j\]so

\[T'(e'_ i)(e_ j) = b_ {ji}\]But by definition of the dual map

\[a_ {ij} = e'_ i(T(e_ j)) = (e'_ i \circ T)(e_ j) = T'(e'_ i)(e_ j) = b_ {ji}\]as required.

Quickly prove that $U$ is mapped isomorphically to $U^{00}$ under the natural isomorphism $\phi : V \to V’’$ given by $\phi(v) = E _ v$ where $E _ v$ is the evaluation map.

Note that $f \in U^{0}$ if $\forall u \in U, f(u) = 0$. By definition, for any $f \in U^{0}$ and any $v \in U$, $E _ v(f) = 0$, so $v \in U \implies v \in U^{00}$. Hence $U \cong E _ v(U) \subseteq U$. Then, using a dimension argument $\dim(U^{00}) = \dim U$, so the result follows.

Suppose:

- $V$ is a finite dimensional vector space
- $\{f _ 1, \cdots, f _ k\} \in V’$
- $\bigcap _ {i} f _ i = \{0\}$

Quickly prove that

\[\langle f_1, \cdots, f_k \rangle = V'\]
(this is in fact an if and only if).

Suppose that $\dim V = n$, and let $L = \langle f _ 1, \cdots, f _ k\rangle$. Mote that $k < n$. Thus

\[\begin{aligned} U &= \bigcap^k_{i = 1} \ker g_i \\\\ &= \ker \phi \end{aligned}\]where we define $\phi : V \to \mathbb F^k$ via

\[\phi(v) = (g_1(v), \cdots, g_k(v))\]Since $k < \dim V$, the rank-nullity theorem applied to $\phi$ gives that $\ker \phi \ne \{0\}$. But since $U = \ker \phi$, this is a contradiction since we are assuming that $U = \{0\}$. So $L = V’$.

(Apparently there is an alternative argument using the natural isomorphism between $V$ and $V’’$, but I can’t figure it out).

What properties need to preserved in an isomorphism of vector spaces?

### Proofs

Suppose $V$ is a finite dimensional vector space with a basis $\{e _ 1, \ldots, e _ n\}$. Show that the dual basis

\[[e_1', \ldots, e_n']\]
where

\[e'_i : e_j \mapsto \begin{cases}
1 &\text{if } i =j \\\\
0 &\text{otherwise}
\end{cases}\]
really is a basis for the dual $V’$.

Todo, linear independent: evaluate linear combination on $e _ j$ for each $e _ j$. Spanning, construct an explicit $f$ formed from dual basis.

Let $T : V \to W$ be a linear map. The dual map $T’$ is defined as

\[T' : W' \to V'\]
where

\[T'f = f \circ T\]
Prove that this map is linear.

Todo.

Let $V$ and $W$ be two finite dimensional vector spaces. Quickly prove that $T \mapsto T’$ defines an isomorphism from $\text{Hom}(V, W)$ to $\text{Hom}(W’, V’)$.

- Need to show that this is an isomorphism of vector spaces, so that $(T + \lambda S)’ = T’ + \lambda S’$
- Then need to show that it is injective, can consider $T’ = 0$
- Then can show surjectivity with standard dimension argument

Prove that

\[v \mapsto E_v\]
where $E _ v : V’ \to \mathbb F \in V’’$

\[E_v(f) = f(v)\]
gives a natural isomorphism between $V$ and $V’’$.

Todo, page 30 of Linear Algebra notes.