$V$ 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?
If the kernel of a mapping is a nullvector, then is the dimension of the kernel is zero, is that right?
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1 Answers
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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.
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0thank you very much for your answer. and if a $dimV = dim(imf_V)$ than it means that $f$ is surjectiv, is that correct? – 2017-01-22
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0Do you perhaps mean that $f$ is *injective*, rather than surjective? Consider the map $f: \Bbb R \to \Bbb R^2$ given by $f(x) = (x, 0)$ (so putting the real line in the Cartesian plane). The image of $f$ does have the same dimension as $V$, but is not surjective. It is injective though, and this follows from the [rank-nullity theorem](https://en.wikipedia.org/wiki/Rank–nullity_theorem), remembering that $f \text{ injective } \iff \ker(f) = \{0\}$. – 2017-01-22
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0and what condition have to be fulfilled to show that f is surjective? – 2017-01-22
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0I guess for $f: V \to W$, just that $\dim ( \operatorname{im} f) = \dim W$; so the image has to have the same dimension as the *range*. Just thinking about the domain, in general, can't tell us about surjectivity. – 2017-01-22