I having trouble understanding the difference between the kernel of a linear transformation and the basis for the kernel of a linear transformation.
My textbook defines the kernel of a linear transformation as
Let $T:V\to W$ be a linear transformation. Then the set of all vectors $v$ in $V$ that satisfy $T(v) = 0$ is called the kernel of $T$ and is denoted by $\ker(T)$.
and later states
… the kernel of $T$ is the solution space of $Ax=0$.
My textbook states in an example that
a basis for the kernel of $T$ was found by solving the homogeneous system
represented by $Ax = 0$.
I don't understand the distinction, if there is any, between being asked to find the kernel of a linear transformation and the basis for the kernel.
Best Answer
Instead of thinking of "the" basis of the kernel, you need to think of "a" basis of the kernel.
The kernel is a subspace of the domain. In general, it doesn't have only one basis; it has many.
For example, consider $T:\mathbb R^3 \to \mathbb R$ given by $T(x,y,z) = x+2y+3z.$ The kernel is the set of all points $(x,y,z)$ for which $x+2y+3z=0.$ If you pick $y$ and $z$ to be any numbers at all and then let $x = -2y-3z,$ then the resulting point $(x,y,z)$ is a member of the kernel of $T.$ The kernel contains infinitely many points because there are infinitely many values of $y$ and $z$ that you could have chosen.
Every basis of the kernel contains only two points, whereas the kernel itself contains infinitely many.
One basis of the kernel is this: $$ \{ (-2, 1, 0),\ (-3,0,1) \}. $$ The first of these points corresponds to the choice $y=1,$ $z=0.$ The second corresponds to $y=0$, $z=1.$
This is a basis for the kernel because every member of the kernel is a linear combination of these two vectors, and this set of two vectors is linearly independent.
Here is another basis of the kernel: $$ \{(2,-1,0),\ (2,0,-1)\}. $$
There are infinitely many different bases of the kernel, and each of them is a finite set, containing only two elements.
There is only one kernel, and it is an infinite set.