Can a linear mapping be invertible when its matrix is not? Let $V$ be a $k$-dimensional subspace of $\mathbb{R}^n$, with $k < n$. Choose an orthonormal basis $(a_1, \ldots, a_k)$ of $V$ and let $A$ be the $k \times n$ matrix whose rows are $(a_1^T, \ldots , a_k^T)$. Define a linear mapping $L: V \rightarrow \mathbb{R}^k$ by
$$
L(x) = Ax.
$$
From the definition of $A$ it is clear that
$$
L^{-1}(y) = A^T y.
$$
Therefore, $L$ has an inverse while $A$ (being rectangular) has not. This seems odd...

While writing this it came to me that $A$ probably isn't the matrix representing $L$ in any basis, since such should be a $k \times k$ matrix, but I am a bit confused - what then is $A$?
 A: It seems people would like me to answer my own question... well then, why not.

My original assumption was that if a linear mapping $L$ can be written as
$$
  L(x) = Ax
$$
for some matrix $A \in \mathbb{R}^{k \times n}$, then $A$ represents $L$ with respect to some basis (in fact, the canonical bases of $\mathbb{R}^n$ and $\mathbb{R}^k$ in this case, as $L$ seems to go from $\mathbb{R}^n$ to $\mathbb{R}^k$).
However, there is a subtelty about the domain of $L$ in the problem as posted originally - here, $L$ goes from a $k$-dimensional subspace of $\mathbb{R}^n$ to $\mathbb{R}^k$. Therefore, both the domain and range of $L$ are actually $k$-dimensional vector spaces, and any matrix representing $L$ will necessarily be $k \times k$.
Now, what does such a matrix look like? To find it, choose bases for the domain $V$ and range $\mathbb{R}^k$ of $L$. For example, $(a_1, \ldots, a_k)$ and $(e_1, \ldots, e_k)$ will do, in which case the matrix will turn out to be the $k \times k$ identity matrix (which is obviously invertible).
Thus, to answer the question in the title, if a linear map is invertible, any matrix representing it will be so as well.
A: I think that a great part of your difficulty was that you were suffering from a mathematical education that hadn’t paid sufficient attention to the importance of the domain, the target space, and the image (=“range”) of a map. Do you know the following about ordinary maps $f\colon X \to Y$ between sets?
The map $f$ is one-to-one if and only if it has a left inverse, i.e. there is a map $g\colon Y \to X$ such that $g\circ f=$ identity on $X$. And $f$ is onto (meaning that the image is equal to all of the target space) if and only if $f$ has a right inverse, i.e. there is a map $h\colon Y \to X$ such that $f\circ h=$ identity on $Y$. The latter is not quite trivial: in fact it depends on (and I think is equivalent to) the Axiom of Choice. The proofs are easy enough, nothing fancy, and if you haven’t seen the facts, you should spend a little time to prove them for yourself.
The significance of the above is that the same results are true for maps between vector spaces. You have constructed a left inverse of the inclusion map $V\to {\mathbb{R}}^n$, which of course is a one-to-one map, but your new map is a left inverse only. Try composing them in the opposite direction!
