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If we have some finite dimensional vector space V and linear transformation $F: V\to V$, and we know that F is injective, we immediately know that it is also bijective (same goes if we know that F is surjective).
I'm curious if a same rule applies if V is infinite dimensional vector space, and we know that F is injective/surjective, does it again immediately imply that F is also bijective (intuitively I think it does)?

Martin Sleziak
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aleksaleks123
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6 Answers6

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No, it does not. Consider the space $\ell_\infty$ of bounded sequences of real numbers. The map

$$S:\ell_\infty\to\ell_\infty:\langle x_0,x_1,x_2,\ldots\rangle\mapsto\langle 0,x_0,x_1,x_2,\ldots\rangle$$

that shifts each sequence one term to the right and adds a leading $0$ term is linear, injective, and clearly not surjective.

Brian M. Scott
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No it does not, take for example the linear operator $D$, the derivative map from

$$C^\infty(\Bbb R)\to C^\infty(\Bbb R)$$

Then this map is clearly surjective by the FTC, but it is not injective as the derivative of any constant is $0$, so the null-space is non-trivial.

Adam Hughes
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  • As a nitpicky comment: this one doesn't exactly address the OP's question (well, not literally) because he asked about operators on the same space $V\to V$. – zipirovich Jan 10 '17 at 01:05
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    @zipirovich fair enough, I'll change the codomain to match the domain, the rest of the result is still true without alteration. Thanks! – Adam Hughes Jan 10 '17 at 01:08
  • In that case you'll have to make your space something like $C^{\infty}(\mathbb{R})$. Otherwise, it's not well-defined because over reals the derivative of a function is not necessarily differentiable again. – zipirovich Jan 10 '17 at 01:13
  • @zipirovich ah yes, of course. Thanks again! – Adam Hughes Jan 10 '17 at 01:14
  • You're welcome! :-) – zipirovich Jan 10 '17 at 01:14
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Another counter-example, taken from algebra:

Let $K$ be a field, and consider the infinite dimensional $K$-vector space of polynomials in one indeterminate, $V=K[X]$. Multiplication by $X$ is a linear map, which is injective, but not surjective, since polynomials with a non-zero constant term are not attained.

Bernard
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It does not.

Take the space of the real sequences, and the transformation

$T(u) = v$ where $v_n = u_{n+1}$

Then $T$ is surjective, but it's not injective.

Tryss
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    Neat, together with Brian M. Scott's answer, we have $TS=\mathrm{id}$ but $ST\ne\mathrm{id}$. –  Jan 10 '17 at 00:48
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Abstract example: take a basis $(b_i)_{i\in I}$ of the space. Let be $J\subset I$ with $|J| = |I|$ and $\sigma:I\longrightarrow J$ a bijection. The linear function defined by $$S(b_i) = b_{\sigma(i)},$$ is injective but not surjective while $$T(b_i) = b_{\sigma^{-1}(i)},i\in J,$$ $$T(b_i) = 0,i\in I\setminus J,$$ is surjective but not injective.

Martín-Blas Pérez Pinilla
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2

Here's another example of a surjective but not bijective map, also on $V=K[X]$, the vector space of polynomials over a field. Taking the formal derivative of a polynomial is a surjective linear map, but not a bijection since for example all constant polynomials are mapped to zero.

zipirovich
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