A bijective linear map with no inverse, how is it possible? In my course I have learned that if $A$ is a matrix $m\times n$ with $m>n$, then $V\to W$ defined by $x\mapsto Ax$ is injective if the columns are linearly independent (with $V$ a vector-space of dimension $m$ et $W$ a vector space of dimension $n$). Therefore it's going to be bijective $V\to \operatorname{span}(\operatorname{clolumn}(A))$ isn't it? But how can this be, since the inverse at left will not be the same as the reverse at right? 
If it's unclear what I'm saying, here is an example. 
Let $A=\begin{pmatrix}1&0\\ 0&1\\ 0&0\end{pmatrix}$. The columns are linearly independant and thus it's injective. Therefore $\mathbb R^2\longrightarrow \operatorname{span}\{(1,0,0),(0,1,0)\}$ defined by $x\mapsto Ax$ is bijective. But its inverse at right is $x\mapsto Bx$ where $B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and the inverse at left is $x\mapsto Cx$ where $C=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$.
But $Cx\neq Bx$, therefore $x\mapsto Ax$ can't be bijective, can it? So how can it be injective and surjective (and thus bijective), and not have an inverse ?
 A: I think you are confusing the dimensions of the spaces you are working with. Lets take a look at your example:


*

*The linear function $x \mapsto Ax$ is a function $A \colon \mathbb{R}^2 \to \mathbb{R}^3$.

*The mapping is injective (into $\mathbb{R}^3$) since the columns of $A$ are linearly independent.

*The function $A$ is surjective into $\operatorname{span}\lbrace (1,0,0), (0,1,0) \rbrace$, as a $2$-dimensional subspace of $\mathbb{R}^3$. Together with (2.) this yields that $A$ is bijective for $\operatorname{span}\lbrace (1,0,0), (0,1,0) \rbrace$.

*However, the function is not surjective with respect to $\mathbb{R}^3$ itself; e.g., $(0,0,1)$ has no preimage, i.e. there is no $x \in \mathbb{R}^2$ such that $Ax = (0,0,1)$. Therefore $A$ is not bijective with respect to $\mathbb{R}^3$.


Note that for linear functions $f \colon V \to W$ the dimensions $V$ and $W$ have to be equal for $f$ to be bijective. However this is not a sufficient condition but a necessary one.
A: Your map is $f:\mathbb{R}^2 \to \operatorname{span[(1,0,0) , (0,1,0)]}$
$$f(x,y) = (x,y,0)$$
which is obviously bijective. Its inverse is simply
$$(x,y,0) \mapsto (x,y)$$
can you figure out what is the matrix associated to it?
