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I'm new to linear algebra and I'm still quite shaky when it comes to all the new notations and concepts.

According to the dimension theorem of vector spaces $\dim(\ker(T)) + \operatorname{rank}(T) = \dim(U)$ for the linear transformation $T:U \mapsto V$. When I was thinking about that and tried to draw the domains and so one I thought about the following: Does the dimension theorem also imply that $\dim(\ker(T)\cup T^{-1}(V)) = \dim(U)$, whereby $T^{-1}$ is the inverse image of the range of $T$?

Johannes Kloos
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MoltenLight
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    The union of subspaces need not be a subspace, see here. So why is $\ker(T)\cup T^{-1}(V)$ a subspace? – Dietrich Burde Oct 09 '20 at 12:48
  • Note that $\ker (T) \subset T^{-1}(V)$ because $0 \in V$. – CyclotomicField Oct 09 '20 at 12:55
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    @DietrichBurde while it's not true in general it is true in this case because the kernel is a subset of the preimage. – CyclotomicField Oct 09 '20 at 12:56
  • @CyclotomicField I was hoping the OP would answer this, but thank you. – Dietrich Burde Oct 09 '20 at 13:07
  • @DietrichBurde Thank you for the link! As I said I'm quite new to linear algebra and I don't understand these concepts all too well yet, but I'm eager to learn. So if $ker(T)$ wouldn't be contained in $T^{-1}(V)$, then $ker(T) \cup T^{-1}$ wouldn't be a subspace. And I guess that in this case $dim(ker(T) \cup T^{-1})$ is not computable? – MoltenLight Oct 11 '20 at 12:37

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$T^{-1}(V)=U$, $\ker T\subseteq U$ hence $\ker(T)\cup T^{-1}(V)=U$ and therefore $\dim(\ker(T)\cup T^{-1}(V)) = \dim(U)$ is true, but the dimension theorem is not needed to prove this.

FormulaWriter
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