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Could you help me solving this problem?

Let $T:V\rightarrow V$ be linear and injective. Prove that if $V$ is finite-dimensional then $T$ is surjective. You are not allowed to use rank nullity theorem.

With the use of R-N theorem the question seems to be very simple, however, I am not really sure how to prove it without using this theorem. I would appreciate any suggestions.

marco11
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    Consider the image of a basis of $V$ – Simon S May 26 '15 at 10:00
  • Can I just prove that $T(e_i)$ for all $i={1,\dots,n}$ is a basis of an image while $e_1,\dots, e_n$ is the basis of $V$? – marco11 May 26 '15 at 10:14
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    Yes, if ${ e_i }$ is a basis of finite dimensional $V$ and $T$ is injective, then ${ T(e_i) }$ is also a basis of $V$. That being the case, it is possible to write any vector $w \in V$ as $w = T(v)$ for some $v \in V$. Can you fill in the details? – Simon S May 26 '15 at 10:21
  • As $Ker(T)={0}$, then $T(e_1),\dots, T(e_n)$ are independent and so they create a basis. So that for every $v \in V$ say, $v=\alpha_1 e_1 +\dots \alpha_n e_n$ and $T(e_i)=\beta_{1i} T(e_1)+\dots+\beta_{ni} T(e_n)$, we can write: $v=\alpha_1 (\beta_{11} T(e_1)+\dots+\beta_{n1} T(e_n)) +\dots \alpha_n (\beta_{1n} T(e_1)+\dots+\beta_{nn} T(e_n))$. Right? – marco11 May 26 '15 at 10:46
  • And the last line simplifies to $\sum^{n}{i=1}\alpha_i \sum^{n}{j=1} \beta_{ji}T(e_j)$ so clearly $T$ is surjective. – marco11 May 26 '15 at 10:53
  • You can probably find several related posts on this site:http://math.stackexchange.com/questions/805625/if-gv-rightarrow-v-is-an-injective-linear-transformation-prove-if-v-is-fi, http://math.stackexchange.com/questions/597463/linear-map-fv-rightarrow-v-injective-longleftrightarrow-surjective, http://math.stackexchange.com/questions/200774/question-on-finite-vector-spaces-injective-surjective-and-if-v-is-not-finite (and probably a few other posts) Maybe also have a look on the list of related question auto-generated on the right. – Martin Sleziak May 26 '15 at 12:59

2 Answers2

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An injective linear map sends linearly independent sets to linearly independent sets. More precisely,

If $T\colon V\to W$ is an injective linear map and $\{v_1,v_2,\dots,v_n\}$ is linearly independent, then $\{T(v_1),T(v_2),\dots,T(v_n)\}$ is linearly independent.

Indeed, if $$ \sum_{k=1}^n \alpha_k T(v_k)=0 $$ then $$ \sum_{k=1}^n T(\alpha_k v_k)=0 $$ and, by injectivity, $\sum_{k=1}^n \alpha_k v_k=0$.

If $\{v_1,v_2,\dots,v_n\}$ is a basis for $V$ and $W=V$, we have that $\{T(v_1),T(v_2),\dots,T(v_n)\}$ is linearly independent in $V$. Then…

egreg
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  • And then we need to show that every $T(v_i)$ can be written as linear combination of $v_i,\dots, v_n$? – marco11 May 26 '15 at 11:02
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    @marco11 You should already know that if a linearly independent set has as many elements as the dimension of the space, then it is a basis. – egreg May 26 '15 at 11:16
  • This is true, so is that enough to conclude that the dimension of domain and codomain is the same? – marco11 May 26 '15 at 11:18
  • @marco11 The domain and the codomain are the same! Now, the range contains a basis of $V$, so the range is $V$. – egreg May 26 '15 at 11:19
  • ah, the codomain and the range I mean, sorry. – marco11 May 26 '15 at 11:19
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Okay well we know that If $T:V\rightarrow W$ is a linear transformation with $dim(V)=dim(W)$ and V and W being finite dimensional vector spaces then following are equivalent:

1) T is non singular(that is kernel is 0)

2) T is invertible

3) T is surjective

We have been given that T is injective,V is finite dimensional that implies T is non singular and from the above mentioned theorem we know that T will be surjective .

Hence proved