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Material covered: some properties of composition of linear maps (products), and an introduction to kernels and images (or nullspaces and ranges).

*1*Composition of linear maps: properties¶

- Associativity
- Existence of an identity, $I \in \mathcal (V, V)$, such that $IT = T$ for any $T \in \mathcal (V, W)$.
- Distributivity, to tie together addition and "multiplication": $(S_1 + S_2)T = S_1 T + S_2 T$

Notably, composition is *not* generally commutative.

*2*Kernels and images¶

Now, onto **kernels** (also known as **nullspaces**) and **images** (also known as **ranges**). (The professor prefers the terms in parentheses, but the notetaker likes the other ones better. Plus, $\LaTeX$ supports \im and \ker but not \null and \range.)

*2.1*Kernels¶

First, let's define the kernel of a transformation. For any $T \in \mathcal L(V, W)$:

$$\ker(T) = \{v \in V \mid T(v) = 0_W \} \subseteq V$$

Some trivial examples: the kernel of the zero map (which just sends everything to 0) is the whole domain, while the kernel of the identity map is just $\{0\}$.

Now we will prove that the kernel of a linear map $T \in \mathcal L(V, W)$ is a subspace of $V$:

- Show that $0 \in \ker(T)$: Well, $T(0) = T(0 + 0) = T(0) + T(0)$ so it must be that $T(0) = 0$, since $T(0)$ behaves as the additive identity (which is just 0).
- Show that scalar multiplication and addition are preserved: let $u,v \in V$ such that $Tv = Tu = 0$, and let $a \in \mathbb F$. Then $T(au + v) = aTu + Tv$ (by the definition of a linear map) $= a\cdot 0 + 0 = 0$. Thus $au + v \in \ker(T)$.

*2.1.1*Relation to injectivity¶

Proposition: If $T \in \mathcal L(V, W)$, then $T$ is injective if and only if $\ker(T) = \{0\}$.

Proof: ($\to$) $T$ is injective. We know that $T0 = 0$, from earlier. So if $Tv = v$ for any $v$, then $v = 0$, by injectivity.

($\rightarrow$) Assume $\ker(T) = \{0\}$. Let $u, v \in V$ such that $T(u) = T(v)$. Then $Tu - Tv = 0$. So $T(u-v) = 0$ by linearity. But this means that $u-v\in \ker(T)$ and so $u-v = 0$. Thus $u = v$.

*2.2*Images¶

Given $T \in \mathcal L(V, W)$:

$$\im(T) = \{Tv \mid v \in V\} \subseteq W$$

If $T$ is subjective, $\im(T) = W$.

Proposition: the image of a linear map is a subspace of $W$. Proof:

- To be continued