Tuesday, January 22, 2013 CC-BY-NC
Kernels and images

Maintainer: admin

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

1Composition of linear maps: properties

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

Notably, composition is not generally commutative.

2Kernels 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.)


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 $V$:

  1. 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).
  2. 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 $au + v \in \ker(T)$.

2.1.1Relation 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 $u = v$.


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:

  1. To be continued