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I just had a linear algebra midterm today and I really dont know how to do the last question. The question goes as follows.

The function $f(x)=x^3+1$ defines an isomorphism between $\mathbb{R}$ and a vector space $V$. The addition and multiplication in $V$ are nonstandard and unknown, while those in $\mathbb{R}$ are just the normal ones. How are the nonstandard ones defined and find the zero vector in $V$.

I really have no idea and I didn't have much time left since this was the last one on the exam. So I just listed values of $f(1)$, $f(2)$, $f(3)$ and $f(4)$ with different ways of getting them. But I couldnt really spot any pattern.

Olivier
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Zeyuan
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  • Hint. Your instructor could have asked this question with any bijection from $\mathbb{R}$ to some set, and used the bijection to define the operations on the image. – Ethan Bolker Mar 30 '17 at 00:37

2 Answers2

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Let's denote the "addition" in V by $\oplus\,$. By definition of isomorphism, $$f(x+y)=f(x)\oplus f(y)$$ for all $x,y\in{\Bbb R}$. Writing $u=f(x)$ and $v=f(y)$ gives $$\eqalign{u\oplus v &=f(f^{-1}(u)+f^{-1}(v))\cr &=f((u-1)^{1/3}+(v-1)^{1/3})\cr &=\bigl((u-1)^{1/3}+(v-1)^{1/3}\bigr)^3+1\ .\cr}$$ I'll leave it up to you to simplify this (if you feel like it) and to do something similar for scalar multiplication.

David
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  • I should have thought about that, why was I so stupid？！ – Zeyuan Mar 30 '17 at 00:42
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    This is called transport of structure or pullback. Seehttp://math.stackexchange.com/a/2133694/589 for instance. – lhf Mar 30 '17 at 00:46
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Well, the words non-standard say it all, so you'd have to think about it this way:

If it is really a vector space isomorphism, you'll get all your operations induced from $\mathbb{R}$ and $f$, for example:

1) The identity element must be $f(0) = 1$ because identity is send to identity.

2) Addition of $f(x)$ and $f(y)$ must be given by $f(x) +_V f(y) := f(x+y)$, e.g. (and this is why this is non-standard) $1+_V 1 = f(0) +_V f(0) = f(0) = 1$ (notice that $1$ is the identity with respect to addition) and $2+_V 2 = f(1) +_V f(1) = f(2) = 9$

3) Similarly, scalar multiplication is given by $\lambda \cdot_V f(x) := f(\lambda x)$, e.g. $\lambda \cdot_V 1 = \lambda \cdot_V f(0) = f(0) = 1$ or $2 \cdot_V 2 = 2 \cdot_V f(1) = f(2 \cdot 1) = f(2) = 9$

The idea is really that f is a bijection and hence you can just carry algebraic structures along it

Sebastian Schulz
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