Is every nonsingular linear transformation invertible? I know that every nonsingular square matrix is invertible.
So every nonsingular linear transformation should also invertible.
But $T\colon\mathbb R^2\to\mathbb R^3$ such that $T(x,y)=(x+y,x-2y,3x+y)$ is nonsingular but inverse does not exist.
So can I conclude that:
"If $T$ is invertible then it is nonsingular but if $T$ is nonsingular then it is not necessarily invertible."
Thanks in advance.
 A: Yes, your conclusion is correct. Every invertible linear map is nonsingular, that is, it kernel is trivial. This is because non-singularity is equivalent to injectivity for linear maps.
However to be invertible a linear map must also be surjective, so non-singularity alone is not enough to conclude that a linear map is invertible. Your example of the function $T$ make this clear.
A: A quick search reveals contradicting conventions.  Here are some well-established terms for properties of a linear transformation $T \colon V \to W$.

*

*$T$ is injective or one-to-one if for all $v \in V$, $T(v) = 0 \implies v=0$.

*$T$ is surjective or onto $W$ if for all $w \in W$ there exists $v \in V$ such that $T(v) = w$.

*$T$ is bijective if $T$ is simultaneously one-to-one and onto $W$.

I found some sources that define nonsingular for linear transformations as equivalent to injective.  However, I am more familiar with the term applied to matrices, and equivalent to “square and invertible”.  So closer to bijective for linear transformations.
I agree with you that the linear transformation you describe is injective but not surjective.
