Dual space

We introduce dual Spaces as an alternate way of understanding inner products. This is a bit esoteric, so feel free to ski.

Linear functionals

A linear functional $\phi$ is a linear map that goes from $\phi: V → F$. It just takes a weighted sum of each component of the vector, so you can think of it as $\phi(v) = a \cdot v$, where $a$ is a vector that defines the linaer functional.

Linear Functionals Define a Dual Space

The vector $a$ is a vector, and because $a$ and $v$ have the same dimension, they are isomorphic.

More formally, for a vector space $V$ over a field F, the dual space $V'$ is the space of linear functionals over this field, denoted as $V' = \mathcal{L}(V, F)$.

Dual Basis Property

The dual basis has a weird property. Say that you had space in $R^3$. Your dual basis can be something like (1, 0, 0), (0, 1, 0), and (0, 0, 1).. When you take the dot product of this basis with the corresponding basis in the non-dual space, you get that $(1, 0, 0) \cdot e_1 = 1$ and anything else is zero. In more formal terms:

The $\varphi$ represents the functional defined by the dual basis, and it "selects" its corresponding non-dual vector.

• Intuition (dot product)

  Let’s try to understand this by looking at the dot product. You can always split a dot product into its basis vectors, like this:

  $$a\cdot x = (a_1e_1 + ... + a_ne_n)\cdot(x_1e_1 + ... + x_ne_n)$$

  The things with the $a$'s are the dual basis, and the things with the $x$ are the normal basis. Note how the dual basis "selects" an element of the non-dual basis, meaning that the distribution in the expanded dot product is very simple.

Dual Maps

A dual map is defined in terms of the dual space. We define the dual map as $T’(\phi) = \phi \circ T$ for all $\phi \in W’$.

So if $T: V → W$, then $T': W' → V'$. Here, $W’$ is the linear functional parameters for $W$, and $V’$ is the linear functional parameters for $V$.

• Proof

  Let’s start with $T$ and we know that $T: V → W$.

  We know that $\phi \in W'$, but the overall $T’(\phi) = \phi \circ T$ operates $V$, so it's in $V'$. Therefore, we conclude that $T'$ maps from $W'$ to $V'$.

Algebraic properties

$(S+T)' = S' + T'$

$(\lambda T)' = \lambda T'$

$(ST)' = T'S'$

• proof: $(ST)'\phi = \phi \circ ST = (\phi \circ S) \circ T = T'(S'(\phi)) = T'S'\phi$

Annihilator

For any subspace $U \subseteq V$, the annilator $U^0$ is defined as the following

This is a subspace in dual $V’$ that has a specific property. We can understand this as the orthogonal complement to $U$, but let’s not use this term just yet.

Annihilator and dual identities

$\{0\}^0 = V'$. Take a second to think about why

$\dim U + \dim U^0 = \dim V$ (makes sense)

$null(T') = (range T)^0$.

• This one is a very important one. It's saying that the "orthogonal complement" of the column space is equivalent to the null space of the dual map. this ties back to the four fundamental subspaces

• Proof

  Let $\phi \in null T'$. Therefore, $T'(\phi) = \phi \circ T = 0$, which means that $\phi \circ T v = 0, v \in V$.

  Therefore, $\phi \circ Tv = \phi(Tv) = 0$, meaning that $\phi$ maps everything in the range of $T$ as zero. This is the definition of an anniliator, so we are done.

  To go the other way, let $\phi \in (range T)^0$. Therefore, $\phi(Tv) = 0$, which means that $T'(\phi) = 0$, meaning that $\phi \in null T'$, as desired.

$range T' = (null T)^0$

• again, this is VERY relevant to our four subspaces! The orthogonal complement to the null space is the row space, which is the range of the dual map

• Proof

  Here we define $T: V → W$ and $T': W' → V'$

  if $\phi \in range T'$, then there exists some $\psi$ such that $\phi = \psi \circ T$. Therefore, $\phi(u) = \psi \circ T(u)$ for all $u$. Select $u\in null T$. Therefore, $\phi(u) = \psi \circ T(u) = 0$. As such, $\phi$ is an annilator of $null T$, and we conclucde that $range T' \subseteq (null T)^0$

  To show that the spaces are equal, you can just show that the dimensions are equal. Note that $\dim range T = \dim V - \dim null T$, and we know that the ranges of the dual and non-duals have the same dimension, so $\dim range T' = \dim V - \dim null T$ . We just need to massage the right hand side now.

  We know that $\dim V = \dim U + \dim U^0$, and let $U = null T$. Our equation becomes the following

  $$\dim range T' = \dim null T^0$$

  as desired.

$\dim null T' = \dim null T + \dim W - \dim V$

• the proof uses $\dim U + \dim U^0 = \dim V$ and the previous lemma.

• it follows that $\dim range T' = \dim range T$

• what does this mean? Well, if we cheat and look at $A^T = T'$, then this theorem states that the row rank and column rank are the same!!

Upshot:

• If $T$ is injective, then the range of $T'$ is everything. Therefore, $T'$ is surjectrive.

• Likewise, if $T$ is surjective then the null of $T'$ is nothing, so $T'$ is injective.

Matrix of the dual map

finally, we will start to tie things together. What does the matrix of the dual map look like?

The matrix of the dual map is just the transpose of the normal map!

• naturally, row rank and column rank theorems come about