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Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$

(i) If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.

(ii) Give example showing (i) is false if $V$ is not finite dimensional.

So on (i), since $V$ is finite dimensional, then $V$ has a basis with finite cardinality and hence $\dim(V)=n$. Also, the following holds, $\dim(V)=\dim(\ker(\alpha) + \dim(\operatorname{im}(\alpha))$ Since $\alpha$ is injective, then the $\ker(\alpha)$ is $0$ which implies that the $\dim(\ker(\alpha)$ is $0$ which implies that $\dim(\operatorname{im}(\alpha))=\dim(V)$ and that shows that $\alpha$ is surjective. (Is that right?)

On the second one, (ii), I just have no idea. I mean shouldn't Axiom of Choice be able to work here?

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What does the axiom of choice have to do with it? (It seems totally random unless you explain...) – rschwieb Sep 22 '12 at 20:52
up vote 3 down vote accepted

The statements (i) and (ii) illustrate that finite-dimensional vector spaces behave in a way similar to finite sets, for a map $X \to X$ on a finite set is injective iff it is surjective. However, if $X$ is an infinite set there are maps that are injective but not surjective (e.g. $\mathbb N \to \mathbb N, x \mapsto x+1$) and vice versa (e.g. $\mathbb N \to \mathbb N, x \mapsto \max\{1,x-1\}$). You can turn this into a counterexample for the corresponding statement for vector spaces: let $V$ be the vector space freely generated by $\mathbb N$, i.e. the vector space of sequences $(a_1,a_2,\ldots) \in k^{\mathbb N}$ where $a_i =0$ for all but finitely many $i$ and consider the linear maps induced by the above maps, i.e. $$(a_1,a_2,\ldots) \mapsto (0,a_1,a_2,\ldots)$$ and $$(a_1,a_2,\ldots) \mapsto (a_1+a_2,a_3,a_4\ldots)$$ respectively. The first map is injective but not surjective, and vice versa for the second map.

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Thank you for your help! – Edgar Aroutiounian Sep 22 '12 at 18:06

The second one does not fail because the axiom of choise, you used the fact the the dimension of the space is finite.

A counter example: $V$ is the space of the the polynomials with coefficients in $\mathbb{R}$ and $T$ is defined by $$T(x^{i})=x^{i+1}$$ (that is $1\mapsto x,x\mapsto x^{2}$ etc' ).

In this case $T$ is clearly $1-1$ but not onto since $1$ is not in the image of $T$

Note $1$: We have that $$dim(Im(T))+1=dim(V)$$ but still $dim(Im(T))=dim(V)$ because both are not not finite, this is what fails.

Note $2$: we defined $T$ on a basis for $V$ so it's well defined.

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I don't think your "note 1" is particularly helpful. It looks like you're trying to invoke the rank-nullity-theorem, but: (a) that is usually only proved for finite dimensions and (b) the nullity of your map is 0, not 1. – us2012 Sep 22 '12 at 19:24
@us2012 - this is not what I tried to explain. since $Im(T)$ doesn't have only the constant poly' which is a subspace of dimension $1$ it follows that the dimension of $Im(T)$ is (in some abuse) one less than the dimension of $Im(V)$...since $-$ is not well defined I turned it to the equation above – Belgi Sep 22 '12 at 19:37

We don't need AC here. Take the following vector space

$$ V:=\left\{\,\{a_n\}_{n=1}^\infty\subset \Bbb R\,\right\} $$

of all the real sequences, with coordinatewise operations (addition and multiplication by scalar), and define

$$\phi:V\to V\,\,,\,\,\phi(\{a_n\}):=\{0,a_1,a_2,...\}$$

Then, it is easy to check that $\,\phi\,$ is a $\,1-1\,$ linear transformation, yet it clearly is not surjective.

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Thank you! I appreciate it. – Edgar Aroutiounian Sep 22 '12 at 18:16

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