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)?
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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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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