Show that this binary operator is linear

Question

Define the binary opration * on $\Bbb R^n$ as $\Bbb R^n \times \Bbb R^n\to \Bbb R$, $(x,y)\mapsto 0$. I wish to prove or disprove that: $\forall x,\ y\in \Bbb R^n, \forall a,\ b\in\Bbb R^n$, both $x\mapsto x*b$ and $y\mapsto a*y$ are linear.

I claim to prove the two are linear. The following is how I approached it:

In order to show both are linear, I need to show
$\forall x,\ y\in \Bbb R^n, \forall a,\ b\in\Bbb R^n$ and $m\in\Bbb R^n$
(i)$(x+y)*b=x*b+y*b$;
(ii)$(mx)*b=m(x*b)$;
(iii)$a*(x+y)=a*x+a*y$; and
(iv)$a*(my)=m(a*y)$.
I did find that (i) through (iv) are all satisfied and their results are all 0.So, is this the way to show the two are linear? Am I on the track?

Answer 1

Yes, this is correct. The result is $0$ all the time, and so there is no way in which the properties could be violated. Both maps are linear. In other words: The map $\ast$ is bilinear.

Answer 2

Yes, the map $*: a* b$ sends each ordered pair $(a, b)$ of $n$-tuples, $(a, b) \in \mathbb R^n\times \mathbb R^n\;$ to the zero n-tuple $\;\bf 0\in \mathbb R^n\;$ is the zero map, which is, by definition linear: each property is necessarily satisfied, as you've learned.

(Also, since the zero vector is necessarily an element of any vector space, by definition, it satisfies the properties of linearity.)

Having shown this binary operation satisfies each of the properties you list is all you need to do to prove linearity.

Interestingly, and you can verify this for yourself, the "zero map", which I suppose we can call the "trivial map", represents the only real-valued $n$-tuple ${\bf x}$ for which there exists a linear map from $\mathbb R^n \to \{\bf x\}$.

Answer 3

Assuming that this indeed is just the zero map. That is you have the map $$ \phi: \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n $$ given by (by definition) $$ \phi(x,y) = x*y = 0 $$ You want to prove that $\phi$ is linear. That is you want to prove that $$ \phi(ax+b,y) = a\phi(x,y) + \phi(b,y)\quad\text{and} \\ \phi(x, ay+b) = a\phi(x,y) + \phi(x,b). $$ But everything is zero, so you clearly have these two requirements satisfied.