In the book "No BS Linear Algebra," the author states that "for linear transformations to be invertible, it only needs to be either injective or subjective," mentioning the invertible matrix theorem where a matrix is invertible iff the matrix's null space is only the 0 vector.
However, a Google search on the Mathematics StackExchange gives a different answer, stating a linear transformation matrix is invertible iff it is bijective, and I'm a bit confused as to which is true.
Best Answer
This is due to the rank-nullity theorem, which states that $$ \dim(Range(T))+\dim(Kernel(T))=n $$ for a transformation $T:\mathbb{R}^n\to\mathbb{R}^n$.
If your transformation is surjective, $\dim(Range(T))=n$ and, therefore, $\dim(Kernel(T))=0$ (it is injective). By the same relation, injective implies surjective. The conclusion is: for linear transformations of a finite-dimensional space, injective, surjective and bijective are synonyms.