Dimensional Vector spoace with injective surjective and bijective

Question

I know there are similar questions on the internet, but I'm not getting it smart. i hope u can help me,

Let $B,C$ finite dimensional Vector space with $\dim(C) = \dim(B)$ and $R \in Hom(B,C)$. Then it should be equivalent to R is injective, R is surjective and R is bijective.

I hope u can help me to construct a proof

Answer 1

As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?

If that is the case, choose a basis for $B$; say, $\lbrace x_1,\dots ,x_n\rbrace$. If $R$ is injective, try showing that $\lbrace R(x_1),\dots ,R(x_n)\rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $\mathrm{span} (R(x_1),\dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.

The proof of surjective $\Rightarrow$ injective is somewhat similar.

Walkthrough. In case you give up, here is a walkthrough of the proof for injective $\Rightarrow$ surjective. Let $\lbrace x_1,\dots ,x_n\rbrace$ be as above. Then $\mathrm{dim} (B) = \mathrm{dim} (C) = n$. We will show that $\lbrace R(x_1),\dots ,R(x_n)\rbrace$ is a basis for $C$.

First we must see that $R(x_1),\dots ,R(x_n)$ are linearly independent. Suppose we have $$ \alpha_1R(x_1)+\dots +\alpha_n R(x_n) = 0. $$ But $R$ is linear, so we get: $$ R(\alpha_1x_1 + \dots + \alpha_nx_n) = 0. $$ Since $R$ is injective, we then have $$ \alpha_1x_1 + \dots + \alpha_n x_n = 0. $$ But $x_1,\dots ,x_n$ are linearly independent, so we must have $\alpha_i = 0$ for each $i=1,\dots ,n$. This shows that $R(x_1),\dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.

Answer 2

Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.

The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:

(i) you should say so!

(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:B\to C$ by $$Rx=(0,x_1,x_2,\dots).$$ Then $R$ is injective but not surjective.

You should give a similar example of a linear map which is surjective but not injective.