I'm having some trouble understanding the Range and Kernel of a linear transformation. The definition goes as follows:
Let $T:V \longrightarrow W$ be a linear transformation. Define the sets $ker(T) = \{v\in V : T(v) =0\}$, $R(T) =\{w\in W : w=T(v)\; for\; some \; v\in V\}$. $ker(T)$ is the kernel and $R(T)$ is the range.
Then this theorem follows:
If $T:V \longrightarrow W$ is linear, then $ker(T)$ is a subspace of V and $R(T)$ is a subspace of $W$.
I'm not really getting it. Could someone please give a simple, yet detailed, explanation along with an example of what the Kernel and Range is and what the theorem that follows wants to say?
Thanks.
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$\begingroup$The definitions are there to highlight sets that are important to understanding the properties of the linear transformation T. Since $T:V\rightarrow W$ the kernel of T is every element of $V$ that T transforms into $0$. The range of T is every element of $W$ that is a transformation of an element of $V$. So, some simple examples:
Let $T:\mathbb{R} \rightarrow \mathbb{R}$ be given by $T(x)=x$. Then Ker$(T)$ = $\{0\}$ (no other element of $\mathbb{R}$ is zero and T is the identity map) and Range$(T)$ is $\mathbb{R}$ because every element of $\mathbb{R}$ is used up by T.
Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by $T(x,y) = (x+y, x-y)$. The kernel here is all elements of $\mathbb{R}^2$ that map to $(0,0)$ under T. This means solving the simultaneous equations $x+y=0$ and $x-y=0$ and you can see that $(0,0)$ is the only solution. So Ker$(T)=\{(0,0)\}$. Range$(T)$ is $\mathbb{R}^2$ again, because if you pick any target point $(\alpha, \beta)$ and solve the simultaneous equations $x+y=\alpha$ and $x-y=\beta$ then you find $x=(1/2)(\alpha+\beta)$ and $y={1/2}(\beta-\alpha)$ , i.e. there is a value (x,y) that T turns into $(\alpha, \beta)$.
What is the theorem telling you? It's telling you that these sets have structure; they're not just random collections of points. In both the above examples the Kernel consists of the origin, and so is a 0-dimensional subspace. If we had an example where the kernel was bigger, it would have to have at least 1 dimension (subspaces have integer dimensions), so it would be a line (or plane, or hyperplane as the number of dimensions increase). In other words, all the elements that T map to zero are related to each other: you can find the line that T maps to zero (that's what the kernel gives you).
As an example here, consider $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y) = (x-y, 0)$. The kernel of T is now $\{(x,y) \in \mathbb{R}^2 : x=y\}$. This is a line in $\mathbb{R}^2$, and T maps any point on it to $(0,0)$.
The range also has structure in the same way (but you expect this because T has structure and T defines the range).
Note also that if the kernel of a linear transformation is just the zero element then the transformation must be injective (one-to-one), which is often very useful to know.
$\endgroup$ 1 $\begingroup$It seems to me that you are not sure how the kernel and range of $T$ are subspaces of $V$ and $W$, respectively. This is all the theorem is saying.
Recall that a subset $U$ of $V$ is a subspace of $V$ if $U$ is a vector space in its own right. We can check that $ker(T)$ is a vector space. If $ker(T)={0}$, we're done. Otherwise $ker(T)$ is nonempty.
Suppose that $V$ and $W$ are vector spaces over a field $F$.
Let $u,v\in ker(T)$. Recall that this means that $T(u)=0$ and $T(v)=0$.
Since $T$ is linear, $T(0)=0$. By extension, $-u\in ker(T)$ because $T(0) = T(u + -u) = T(u) + T(-u) = 0+T(-u)=T(-u)=0.$ Thus the kernel contains additive inverses.
Furthermore, $T(u+v)=T(v)+T(v)=0+0=0$, so $u+v \in ker(T)$.
Finally, for all $a\in F$, $T(au)=aT(u)=a\cdot 0=0$. Thus the kernel is closed under scalar products.
Since the kernel contains 0, contains additive inverses for each element, is closed under addition and scalar multiplication, it is a subspace of $V$. A similar argument can be made for the range of $T$.
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