0
$\begingroup$

Show that $(\mathbb{R}, \mathbb{R}, \oplus , \odot )$ is a vector space if $\oplus$ and $\odot$ are defined by

enter image description here

for all vectors $x, y \in \mathbb{R}$ and scalars $\alpha \in \mathbb{R}$.

I have to show this, but I don't even know where to start

$\endgroup$
2
  • $\begingroup$ What's the definition of a vector space? $\endgroup$ – mizh Dec 16 '19 at 11:42
  • $\begingroup$ For example, one of the properties of a vector space is that $(x \oplus y) \oplus z = x \oplus (y \oplus z)$. Using these definitions, can you verify that that is true? $\endgroup$ – Mees de Vries Dec 16 '19 at 12:03
3
$\begingroup$

This is a typical case of transport of structure via bijections. In general, assume you have an arbitrary field $K$, a $K$-vector space $V$ and a permutation $f$ of $V$ (i.e. a bijection from $V$ to itself); via this permutation you can define two new operations, an internal one on $V$ given by $$\oplus: V \times V \to V, x\oplus y=f(f^{-1}(x)+f^{-1}(y))$$ and an external one $$\bullet: K \times V \to V, \lambda \bullet x=f(\lambda f^{-1}(x))$$

Then it is an easy verification that the new structure $(V, \oplus, \bullet)$ is also a $K$-vector space such that $f$ implements an isomorphism between the original structure on $V$ and the new one; this new structure is said to be transported from the old one via $f$.

In your particular case the original structure is that of $\mathbb{R}$ as a left vector space over itself and the permutation considered is $$f: \mathbb{R} \to \mathbb{R}, f(x)=\sqrt[5]{x+7}$$

$\endgroup$
1
  • 1
    $\begingroup$ This answer is perfectly correct, and this is a good perspective, but in this case it might also be useful for OP just to do the grunt work of verifying the vector space axioms directly. Doing that is definitely an understanding to have, and from their comments it seems that they currently do not. $\endgroup$ – Mees de Vries Dec 16 '19 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.