5
$\begingroup$

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!

$\endgroup$
2
  • 3
    $\begingroup$ Right, you need an infinite-dimensional space. Can you think of some simple infinite dimensional vector space? $\endgroup$
    – Daniel Fischer
    May 24, 2015 at 22:16
  • $\begingroup$ Just simple integration of functions will probably do so we will never obtain a constant, right? $\endgroup$
    – marco11
    May 24, 2015 at 23:12

2 Answers 2

Reset to default
8
$\begingroup$

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 $g$

$\endgroup$
2
  • $\begingroup$ Thank you! Does it mean that it is not surjective? Isn't it not injective? $\endgroup$
    – marco11
    May 24, 2015 at 22:22
  • 2
    $\begingroup$ @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 $). $\endgroup$
    – hmakholm left over Monica
    May 24, 2015 at 22:42
5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Pick your favorite vector space of infinite countable dimension for that. $\endgroup$
    – Daniel Valenzuela
    May 24, 2015 at 22:19
  • 1
    $\begingroup$ I thought it worth mentioning this is the right shift operator. $\endgroup$
    – Squirtle
    Jul 19, 2015 at 4:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .