Proving that $\dim T^{-1}(E) = \dim(\operatorname{Ker}T) + \dim (\operatorname{Ker}T\cap\operatorname{Im}(T))$

Question

$\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\Im}{Im}$ My work: I know that $\dim(T^{-1}(E)) = \dim(\Ker T^{-1}(E)) + \dim(\Im(T^{-1}(E)))$.

I was able to show that $\Ker(T^{-1}(E)) \subseteq \Ker(T)$ and $\Ker (T)\subseteq \Ker (T^{-1}(E))$ and so $\Ker(T^{-1}(E))= \Ker(T)$ but I don't know how to go ahead with the second bit. I want to show that $\Ker( T) \cap \Im(T) = \Im(T^{-1}(E))$. Is this the right approach?

Answer 1

Hint:

To prove that $\; \dim T^{-1}(E) = \dim(\operatorname{Ker}T) + \dim (E\cap\operatorname{Im}(T))$, consider the restriction of the linear map $T$ to $T^{-1}(E)$, which may be denoted $T_E$, and apply the rank nullity formula to this restriction. You'll have to show that

• $\operatorname{Ker}(T_E)=\operatorname{Ker}(T)$;
• $\operatorname{Im}(T_E)=E\cap\operatorname{Im}(T)$.