My intuition on invertible linear transformation $A:V \rightarrow W$ is that there is another linear transformation that kind of sends the vectors back so that the composition of two functions do nothing ($I$: Identity matrix).
If this is the case (Obviously wrong), then we only need left invertible $B:W \rightarrow V$ and $BA = I$.
Why we need the linear transformation be both right and left invertible?
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$\begingroup$I'll use the following definition: To say that a function $A:V \to W$ is invertible means that $A$ is both one-to-one and onto. It's a fact that $A$ is one-to-one if and only if $A$ has a left inverse, and $A$ is onto if and only if $A$ has a right inverse. So, $A$ is invertible if and only if $A$ has both a left inverse and a right inverse.
Let's prove that $A$ is onto if and only if $A$ has a right inverse. First, suppose that $A$ has a right inverse $B: W \to V$, so that $AB = I_W$. If $w \in W$, then $A(B w) = (AB)w = w$. This shows that $A$ is onto.
Now, to prove the other direction, suppose that $A$ is onto. If $w \in W$, there exists $v \in V$ such that $Av = w$, so the set $S_w = \{ v \in V \mid Av = w\}$ is non-empty. By the axiom of choice, there exists a function $B: V \to W$ such that $B w \in S_w$ for all $w \in W$. This function $B$ has the property that $A B = I$. This shows that $A$ has a right inverse.
$\endgroup$ 0 $\begingroup$Think in terms of dimension of these vector spaces. In order for $A$ to be invertible you necessarily need $dim(V)=dim(W)$. Moreover, if $A$ is surjective but not injective so $dim(V)>dim(W)$ and you can define a linear transformation $B$ in a way that $BA=I$ (for this it suffices take a basis of $W$ and define $B$ sending this basis "back" to $V$ via $A$).
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