I have a homework question that ask to prove two parts and I did one part and not sure how to prove the second part. The question says: Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be two normed vector spaces and assume that $T \in L(X,Y)$ is an isomorphism. Define a scalar valued function $\|\cdot\|_T$ on $X$ by $\|x\|_T = \|T(x)\|_Y$ for all $x \in X$. Prove that $\|\cdot\|_T$ defines a norm and show that it is equivalent to the original norm $\|\cdot\|_X$.
As I mentioned, I was able to prove the four axioms of the norm, but for the second part that $\|\cdot\|_T$ and $\|\cdot\|_X$ are equivalent, we need to show that there are $m, M >0$ such that $m\|x\|_X \leq \|x\|_T \leq M\|x\|_X$ for all $x \in X$. I am not sure how exactly to start with this inequality.
In Functional Analysis an isomorphism of normed linear spaces is a bijective map $T$ such that $T$ and its inverse are both continuous. So we have $\|Tx\| \leq M\|x\|$ where $M=\|T\|$ (the operator norm). Also, $\|x\| =\|T^{-1}(TX)\| \leq \|T^{-1}\| \|Tx\|$ so $\|Tx\| \geq m\|x\|$ where $m=\frac 1 {\|T^{-1}\|}$.