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Proof that every linear map $\phi: \mathbb{R} \to \mathbb{R}$ is bijective or zero.

It's not true for $\mathbb{R}^n, n \geq 2$, but how to proof/argue that it is true for $\mathbb{R} \to \mathbb{R}$?

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Do you know what a linear map $\mathbb R\to\mathbb R$ looks like? – Jonas Meyer Feb 2 '12 at 21:53
$\mathbb{R}$-linear? – Arturo Magidin Feb 2 '12 at 21:56
up vote 6 down vote accepted

If by "linear", you mean $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(cx)=c\,\phi(x)$, then argue as follows:

If $\phi(1)=a $, then $\phi(b) =\phi(b\cdot1) =b\phi( 1) =ba$.

If $a=0$, then $\phi$ is identically 0. If $a\ne0$, then, as easily verified, $\phi$ is bijective.

This isn't true if we only require that $\phi(x+y)=\phi(x)+\phi(y)$. Let $\{r_\alpha\}$ be a Hamel basis of $\Bbb R$ over $\Bbb Q$. Let $\phi$ map $x$ to $c_{\alpha_1}+\cdots+c_{\alpha_k}$, where the (unique) basis representation of $x$ is $c_{\alpha_1}r_{\alpha_1}+\cdots+c_{\alpha_k}r_{\alpha_k}$. Then $\phi(x+y)=\phi(x)+\phi(y)$, but takes on only rational values.

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The kernel of an $\mathbb{R}$-linear map $\mathbb{R}\to \mathbb{R}$ is a subspace of $\mathbb{R}$, hence it is $\mathbb{R}$ or $\{0\}$, since these are the only ones.

If it is $\mathbb{R}$, this means the map is zero and we have finished.

If it is $\{0\}$, this means the map is injective.

In this case, we can conclude in two different ways:

1) The image of the map is also a subspace of $\mathbb{R}$. Hence, it is $\{0\}$ or $\mathbb{R}$. It can't be zero, for you can't have zero image and zero kernel. Therefore it must be $\mathbb{R}$, meaning it is surjective. Hence, it is bijective.

2) An injective linear map between linear spaces of the same dimension is automatically bijective.

The second approach is more immediate, but I thought of giving the first one, since it is, mutatis mutandis, the proof of Schur's lemma.

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