Answer to Question #286449 in Linear Algebra for Calvin

Question

. Define the kernel and the image of a linear transformation M : V → W and hence Show that the Kernel and image of M are vector subspaces of V and W respectively. 

Answer 1

Solution:

Let V and W be vector spaces and let $T: V \rightarrow W$ be a linear transformation. Then the image of T denoted as im (T) is defined to be the set

$\{T(\vec{v}): \vec{v} \in V\}$

In words, it consists of all vectors in W which equal $T(\vec{v})$ for some $\vec{v} \in V$ . The kernel, $\operatorname{ker}(T)$ , consists of all $\vec{v} \in V$ such that $T(\vec{v})=\overrightarrow{0}$ . That is,

$\operatorname{ker}(T)=\{\vec{v} \in V: T(\vec{v})=\overrightarrow{0}\}$

Then in fact, both $im (T)$ and $\operatorname{ker}(T)$ are subspaces of W and V respectively.

Statement:

Let V, W be vector spaces and let $T: V \rightarrow W$ be a linear transformation. Then $\operatorname{ker}(T) \subseteq V and \operatorname{im}(T) \subseteq W$ . In fact, they are both subspaces.

Proof:

First consider $\operatorname{ker}(T)$ . It is necessary to show that if $\vec{v}_{1}, \vec{v}_{2}$ are vectors in

$\operatorname{ker}(T)$ and if a, b are scalars, then a $\vec{v}_{1}+b \vec{v}_{2}$ is also in $\operatorname{ker}(T)$ . But

$T\left(a \vec{v}_{1}+b \vec{v}_{2}\right)=a T\left(\vec{v}_{1}\right)+b T\left(\vec{v}_{2}\right)=a \overrightarrow{0}+b \overrightarrow{0}=\overrightarrow{0}$

Thus $\operatorname{ker}(T)$ is a subspace of V.

Next suppose $T\left(\vec{v}_{1}\right), T\left(\vec{v}_{2}\right)$ are two vectors in im (T). Then if a, b are scalars,

$a T\left(\vec{v}_{2}\right)+b T\left(\vec{v}_{2}\right)=T\left(a \vec{v}_{1}+b \vec{v}_{2}\right)$

and this last vector is in $\operatorname{im}(T)$ by definition.