Endomorphism $T : V \to V$ $ \Leftrightarrow \det T \neq 0$ Prove that $T$ is bijective $\Leftrightarrow \det T \neq 0$, where $T$ is a linear map $T : V \to V$ of a finite-dimensional vector space $V$.

Note that the determinant of the square matrix $A$ of the linear map $T$ is denoted by $\det T$ because $\det A$ depends only on $T$.(Not sure if this is common notation or not so I put this here)



I'm not sure what I can do to prove this. I don't think I've yet developed an intuition for linear algebra but I sadly don't see the connection between the two facts we need to prove.
Recall three  facts:

1) Let $A$ be a square matrix, and let $B$ be another matrix, obtained from $A$ via an elementary operation. Then $\det A=0\Leftrightarrow\det B=0.$

2) If the rows of $A$ are linearly dependent, then it is possible to obtain from $A$ a matrix with a $0$ row via elementary operations.

3) If $A$ represents a transformation which is not bijective, then the rows of $A$ are linearly dependent.

On the other hand, if $A$ represents a bijective transformation, then $A$ is invertible. By multiplicativity of the determinant we have$$\det A\cdot\det(A^{-1})=1,$$which yields$$\det A\neq0.$$
If $A$ is invertible then $AA^{-1} = I$ and then $\det(A) \det(A^{-1}) = 1$ and then $\det(A)$ is invertible and therefore not zero. 

Now, since $A A' = A'A = \det(A) I$ where $A'$ is the classical adjoint of $A$, $\det(A)\neq0$, implies that $\frac{1}{\det(A)} A'$ is the inverse of $A$