If the kernel of a mapping is a nullvector, then is the dimension of the kernel is zero, is that right?

Question

$V$ and $W$ are vector spaces. There is a mapping $f: V \to W$ and the kernel of $f$ is trivial, then $\dim(\operatorname{kern} f)=0$, independently of the dimension of $V$, right?

Answer 1

Yes, this is correct: The (sub)space $\{0\}$ always has dimension $0$, because dimension is measured in an absolute way, by asking for the cardinality of a basis. That cardinality doesn't change depending on the ambient space in which $\{0\}$ lives.

Although dimension does not, there is a concept that measures the relative dimension of a subspace $U \le V$ called the codimension of $U$ in $V$. It measures the difference in their dimensions, so that $\operatorname{codim}_V(U) = \dim V - \dim U$. In this case, the subspace $\{0\} \le V$ always has codimension $\operatorname{codim}_V(0) = \dim V - \dim(\{ 0 \}) = \dim V$. But, again, that's not what's being measured here.