Show dim(Ker$(T)) = $dim(Ker$(QTP))$ Let $P: V \to V$ be an isomorphism of $V$ and $Q: W \to W $ an isomorphism of $W$. Let $T: V \to W$ be a linear transformation. 
Prove dim(Ker$(T)) =$ dim(Ker$(QTP))$. 
Attempt: 
Suppose $0 = T(P(v)) \to Q(0) = 0 = Q(T(v))$  Then clearly Ker$(T)$ is a subset of Ker$(QTP)$. But since $Q$ is an isomorphism, we have Ker$(QTP) = 0.$ Hence since the kernel must be at least of dimension $0$, we have Ker$(T) = 0$. The claim follows. 
I'm not sure if this is correct.
 A: We know that $Q$ is an isomorphism, which implies that $\ker(Q) = \{0\}$.  Now, let $v \in \ker(QTP)$.  This means $Q(T(P(v))) = 0$; but since the kernel of $Q$ is trivial, this implies that $T(P(v)) = 0$, i.e. $v \in \ker(TP).$  We therefore have $\ker(QTP) \subseteq \ker(TP)$, and the reverse inclusion is trivial, so we have $\ker(QTP) = \ker(TP).$ 
Now, note that $P^{-1} \left(\ker(T) \right) = \ker(TP)$.  Since $P$ is an isomorphism, $P^{-1}$ is as well, so $\dim(\ker(T)) = \dim(P^{-1}(\ker(T))) = \dim(\ker(TP))$. 
A: I'm going to assume you're working with finite dimensional vector spaces.
Using, rank-nullity, we can consider the range of $QTP$. Since $P$ is an isomorphism, it follows that the range of $TP$ will be equal to the range of $T$. Then, since $Q$ is also an isomorphism, the kernel of $Q$ is $\{0\}$, so the dimension of the range of $QT$ will be equal to the dimension of the range of $T$. 
We have that the dimension of the range of $QTP$ is equal to the dimension of the range of $T$, so the the claim follows from rank-nullity.
