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Let $ V $ be a vector space of $ \mathbb R$ over the field $\mathbb Q$ and $x$ is an irrational number. I want to prove or disprove that there is a linear function $f$ from $ V $ to $ V $ that satisfies $f(1) = 1 $ and $f(x) = -1$. Since $1$ and $x$ are independent vectors in $V$ I tend to think that such function exists but could not find a way to prove or disprove it.

Tomer
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    You can prove it only if you assume the axiom of choice, I think. – Trebor Sep 11 '21 at 06:41
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    Any independent set can be extended to a basis (assuming the axiom of choice). Then, any function defined on a basis extends uniquely to a linear function. – Berci Sep 11 '21 at 06:49
  • Even if the dimension of the basis is infinity ? – Tomer Sep 11 '21 at 06:53
  • A basis doesn't have a dimension, Tomer. But, yes, even if the basis is infinite, even if the vector space is infinite-dimensional, what @Berci wrote holds true. – Gerry Myerson Sep 11 '21 at 23:53
  • See also https://math.stackexchange.com/questions/4247224/considering-reals-as-vectors-over-rationals-can-i-have-a-linear-map-that-has-un and the several questions linked there. – Gerry Myerson Sep 11 '21 at 23:55

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Assuming the axiom of choice it is shown that every vector space has a basis. Since $1,x$ are linearly independent we can extend it to a basis $B$ of $V$.

Taking hints from comments of Berci, define a function $f$ on $V$ such that $f(1)=1, f(x) = -1$ and $f(v) =0$ for all other $v \in B$ and $$f(y) = \sum_{v_i \in B}a_if(v_i) $$ where $y=\sum_{v_i \in B}a_iv_i$ and $a_i \in \mathbb{Q}$. Note that for a given $y$ , $a_i$ are unique so the function $f$ is well-defined. I let you verify that $f$ is a linear function.

Infinity_hunter
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  • Can you define $f$ to be also one to one ? – Tomer Sep 11 '21 at 07:15
  • Yes, we can surely show the existence of such a map. We consider a map $g$ such that $g(1)=1,g(x)=-1, g(0)=0$ and we can extend this map to a bijection from $B$ to $R$ since the basis $B$ must have a same cardinality of $\mathbb{R}$. – Infinity_hunter Sep 11 '21 at 07:27
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    @Tomer No, $f$ cannot be injective $\mathbb{R}\to\mathbb{R}$: $f(x)=-1$ and $f(1)=1$ means $f(1+x)=0=f(0)$ but obviously $x\neq -1$. – user10354138 Sep 11 '21 at 08:10
  • I see but I think I can make sure that (1+x) is the only basis vector for the kernel of f. – Tomer Sep 11 '21 at 19:07
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    @Tomer yes, just change to $f(v)=v$ for $v\in B$ other than $x$, for example. – user10354138 Sep 12 '21 at 10:12