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Let $W$ be a finite dimensional vector space. $T$ is a linear transformation from $W$ into $W$ and $B$ is a subspace of $W$.

$T^{-1}\left(B\right)\:=\:\left\{v\:∈\:W\left|T\left(v\right)\:∈\:B\right|\right\}$

Prove that $dim\left(T^{-1}\left(B\right)\right)=dim\left(Ker\left(T\right)\right)+dim\left(B\:∩\:Im\left(T\right)\right)$

Test this result on $W=P3,\:T\left(p\right)\left(x\right)=p'\left(x\right),\:B\:=\:P1$

I understand that by the rank nullity theorem, the dimension of this space would simply be the sum of the dimension of the image and kernel. So by this, it seems that $Im(T)$ $∩$ $B$ is equal to $Im(T^-1(B))$ and $Ker(T)$ is equal to $Ker(T^-1(B))$. So I was just wondering if stating these facts is enough of a proof or do I actually need to find out why these rules apply? If so, how exactly would I do that?

And to test the result, would it simply be the case that $P3$ has dimension $4$ and its derivative has dimension $3$ and $P1$ has dimension $2$? Because this gave me that the dimension of $B$$p'(x)$ was $2$ (with $p'(x)$ representing the image of $T$), the kernel of $T$ had dimension $1$, and the dimension of $T^-1(B)$ was $3$. So the formula held as $3$ $=$ $1$ $+$ $2$.

If anyone can provide any feedback on what I have done, I would greatly appreciate it!

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Since $B$ is a subspace of $W$, $T^{-1}(B)$ is a subspace of $W$. Now consider the linear map $T$ restricted to $T^{-1}(B)$ and apply rank-nullity theorem to the linear map $T|_{T^{-1}(B)} : T^{-1}(B) \rightarrow W$. Therefore you will have

$dim(T^{-1}(B)) = rank (T|_{T^{-1}(B)}) + nullity (T|_{T^{-1}(B)})$

You can see that $nullity (T|_{T^{-1}(B)})=nullity(T)$ and $ rank (T|_{T^{-1}(B)}) = dim (B \cap Im(T))$

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