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Let $V,W$ be a vector spaces of the same dimension $m$ and $f\colon V\to W$ be a linear map.

I know that for finite $m$, $f$ is injective $\Leftrightarrow$ $f$ is surjective $\Leftrightarrow$ $f$ is isomorphism.

Is it true for infinite $m$?

Martin Sleziak
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Canis Lupus
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  • No. Just like you have non-surjective but injective mappings of the natural numbers and vice versa. – mkl Jun 13 '13 at 16:29

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Well, no. Take $V = W = k^\mathbb{N}$, which we read as the $k$ sequences indexed by $n$ with $n$ ranging over all of $\mathbb{N}$. Then define the shift operator

$$ S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$$

The map is clearly injective, and equally clearly not surjective.

citedcorpse
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As a counterpart to exitingcorpse's injective-not-surjective map, I'd also like to add $T(x_1,x_2,x_3\dots):=(x_2,x_3,\dots)$ as a surjective-not-injective map.

In fact, you can easily see that $S(T(x))=x$ for all $x$, showing that $ST$ is the identity map, but $TS$ is not the identity map! It's a useful example to have in your pocket...

rschwieb
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    exactly. i'm forever cursed that the first person to show me these examples made typewriter motions with his hands whenever he discussed them, and i have inherited this trait – citedcorpse Jun 13 '13 at 16:32
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    @exitingcorpse What are you doing here then!? Almost everybody is making typewriter motions while answering here :) – rschwieb Jun 13 '13 at 16:34
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No. Let $V=l_2$ (this is the space of sequences $(x_1,x_2,\ldots)$ of complex numbers such that $\Sigma_i |x_i|^2$ is finite), $S:V\to V$ $S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)$. $S$ is injective, but not surjective. See this answer.

Community
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dc2814
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No.

$\varphi:\begin{array} \Bbb R \left[X\right] \to \Bbb R\left[X\right]\\P \mapsto P'\end{array}$

$\Bbb R \left[X\right]$ represents the set of polynomials with coefficients in $\Bbb R$. $P'$ is the formal derivative of $P$.

It's clearly surjective but not injective since all constant polynomials have $\Bbb 0$ as image.

xavierm02
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