My book doesn't cover the criterion for bijective transformations very well. I just want to check my understanding: is this statement true?
Let F be a linear transformation. Let A be the matrix that represents that transformation (which means that that $F(v)=Av$ for any vector $v$). We now have that F is bijective iff $\det(A)\not=0$.
I think about this in the following manner. A bijective linear transformation should have an inverse. Hence the associated matrix should also be invertible. Therefore it's determinant is non-zero. Hope this helps.
Suppose $A$ is invertible. Then $1=\det I=\det(AA^{-1})=\det(A)\det(A^{-1})$. Conversely, if $\det(A)\neq0$ then $\ker(A)=\{0\}$. Because if not, there exists a vector $v\neq0$ s.t. $Av=0\cdot v$, i.e., $0$ would be an eigenvector of $A$, hence $\det(A)=\det(A-0\cdot I)=0$.
