# examples of linear map $f:V \rightarrow V$, which is injective but not surjective

I am trying to find a linear map $f:V \rightarrow V$, which is injective but not surjective.

I always thought that if the dimension of the domain and codomain are equal and the map is injective it implies that a map is surjective. Maybe we need an infinite basis of the vector space $V$? What can be an example of that?

Thank you!

• Right, you need an infinite-dimensional space. Can you think of some simple infinite dimensional vector space? May 24, 2015 at 22:16
• Just simple integration of functions will probably do so we will never obtain a constant, right? May 24, 2015 at 23:12

Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$
• @marco11: $f$ is injective, because if $g_1$ and $g_2$ differ at some point -- say $g_1(x_0)\ne g_2(x_0)$, then $f(g_1)$ and $f(g_2)$ will have different derivatives at that point, and so cannot be the same function. On the other hand $f$ is not surjective because, for example, the function $x\mapsto x+1$ is not $f(g)$ for any $g$ (namely $f(g)(0)=\int_0^0 g(t)dt = 0 \ne 0+1$). May 24, 2015 at 22:42
In finite dimensions we have that bijectivity $\Leftrightarrow$ injectivity $\Leftrightarrow$ surjectivity. Hence we have to come up with an infinite-dimensional example. The idea is to pick a basis $v_i, i\in \mathbb N$, and shift every basis vector $v_i \mapsto v_{i+1}$. We can do that not hitting the first basis vector only because we have infinitely many elements.