Which of the given linear maps are scalar multiplication

Question

The question is:

Which of these linear maps $(v,u): R^2\to R$ is a scalar multiplication. Provide arguments for your answer.

1. $(v, u) = ((x_1, y_1),(x_2, y_2)) = x_1\cdot y_1 + x_2\cdot y_2$
2. $(v, u) = ((x_1, y_1),(x_2, y_2)) = x_1\cdot x_2-y_1\cdot y_2$

How linear map is related to the scalar multiplication? As far as I see it, none of these linear maps produce scalar multiplication, since it has to be:

$(v, u) = ((x_1, y_1),(x_2, y_2)) = x_1\cdot x_2+y_1\cdot y_2$

but I think I am wrong here.

Answer 1

Answer to your question depends on the definition of what you call scalar multiplication. Let me suggest few possible definitions and give corresponding answers.

Mostly, by scalar multiplication we mean endomorphism of some vector space given by multiplying an element of this space by some number (this number is taken from the number field over which the space is defined). Say we take $\mathbb{R}^n$. In this definition then, the scalar multiplication looks like $\lambda \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$, where $\lambda \in \mathbb{R}$, and the map acts on vectors as $$\lambda (u_1, u_2, \dots, u_n) = ( \lambda u_1, \lambda u_2, \dots, \lambda u_n) \ . $$ You see that this is something very different from the scalar multiplication you are talking about (I guess). If not, the answer to your question is: neither of 1. or 2. is scalar multiplication, since the domain and codomain of the map does not have the same dimension.

If the scalar multiplication stands for the inner product, then the question makes a better sense but I have never seen the inner product be called scalar multiplication. Synonyms for the inner product may be scalar product or dot product. This names refers to a positive-definite symmetric bilinear form. If we fix our attention to the case of $\mathbb{R}^n$), this inner/scalar product thing is a map $f \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ which satisfies certain axioms (see this wiki article for their precise form). If we use this definition for your "scalar multiplication" then neither one of the given maps is a scalar product. It satisfies bilinearity as well as the symmetry but fails on positive-definiteness since for $u = (0,1)$ we obtain $$ ((0,1),(0,1)) = -1 \ .$$ The last thing sensible to me to consider is the definition of a pseudo-inner product which keeps all the axioms of an inner product but the axiom of positive-definiteness. The second of your maps actually is a pseudo-inner product (the first is not since it fails to be bilinear). Nevertheless, this last option seems to me the most improbable because calling a pseudo-inner product a scalar multiplication would be completely missleading.

I hope a provided at least some answers and hints for you to be able to solve the problem.

Answer 2

Ok, so I learned from my algebra theacher that scalar multiplication, or scalar product (might be a translation issue, since I am translating from my native language (Lithuanian)) here is meant by such a map between two objects $(.,.): V*V\to R$, where $V$ is a vector space, that satisfies the following five conditions:

1. $(v_1+v_2, u) = (v_1,u) + (v_2,u) , \forall v_1,v_2,u \in V$
2. $(v,u) = (u,v), \forall u, v \in V $
3. $(av, u) = a(v,u)$
4. $(v,v) \ge 0, \forall v \in V $
5. $(v,v) = 0 \iff v = 0 $ (zeroeth element)

What remains is just to test the two parts of the exercise against these conditions.

It seems that the vector space with such scalar multiplication is called Euclidean space.

P.S. Do not hesitate to correct me here, I've just written what I have from my algebra exercise book. There might be discrepancies between terms and definitions.