kernel vs basis for the kernel 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.
 A: 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.
A: If $T$ be a linear transformation from the vector space $V$ to $U$ then kernel of linear transformation $T,$ denoted by $\ker(T)$ is a subspace of $V$. Thus $0\in \ker(T)$. But when we talk about the basis, say $B$ of $\ker(t)$, we look for the maximal linearly independent set $B$ of $V$ which generates $\ker(T)$. You may observe that $0\in \ker(T)$ but not in basis $B$ of $\ker(T)$.
A: For a linear transformation $T$, the kernel is a linear space by itself. Indeed


*

*$T 0 = 0 $

*If $u,v \in \ker(T)$ then $T(a u + b v) = a Tu + b Tv = 0$, therefore $a u + bv \in \ker(T)$


You can then define a basis for this space, which will be basis for the kernel
A: The kernel of a linear transformation is a subspace of the domain of that linear transformation, while a basis for the kernel of a linear transformation is a spanning linearly independent set for the kernel.
If $K$ is the kernel of a linear transformation and $B$ is a basis of it, then in particular we have $B\subseteq K$, so that a basis of the kernel is found first by solving the homogeneous linear system $Ax=0$ and then making some wise choices. But $B\neq K$.
