Suppose $A$ is a bijective linear mapping and $Y$ is a Banach space.

Let X be a vector space and Y a normed subspace with norm $$||_Y$$. Let $$A : X \rightarrow Y$$ be a linear mapping.

(a) Prove that with the prescription $$||x|| = ||Ax||_Y$$ for every $$x \in X$$, a norm is defined on the space $$X$$ if and only if $$A$$ is an injective mapping.

(b) Suppose $$A$$ is a bijective linear mapping and $$Y$$ is a Banach space. Prove that in this case, the space $$X$$ with the norm defined in part (a) becomes a Banach space as well.

I was able to prove (a), but not (b). However, my attempt:

(b) Now, let's assume $$A$$ is a bijective linear mapping and $$Y$$ is a Banach space. To show that $$X$$ with the norm defined in part (a) becomes a Banach space, we need to show that every Cauchy sequence in $$X$$ converges to a limit in $$X$$.

Let $$(x_n)$$ be a Cauchy sequence in $$X$$. Since $$X$$ is a normed space, a Cauchy sequence is also bounded. Since $$A$$ is bijective, its inverse $$A^{-1}$$ exists.

Define $$y_n = Ax_n$$. Since $$Y$$ is a Banach space, every Cauchy sequence in $$Y$$ converges to a limit in $$Y$$. Thus, $$(y_n)$$ converges to some $$y \in Y$$.

How to continue?

Ok, you got that $$Ax_n\to y$$. Now write $$y=Ax$$ for $$x\in X$$. By the definition of the norm on $$X$$:
$$||x_n-x||=||A(x_n-x)||_Y=||Ax_n-Ax||_Y=||Ax_n-y||_Y\to 0$$
And so $$x_n\to x$$.