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I remember when I took linear algebra that there are myriads of vector spaces with myriads of different inner products. But there is one thing that was never clear to me: Given a crazy vector space, how do we construct an inner product or prove that we can't have an inner product defined on it?

This question came up to me because I remembered the following vetor space:

$$\oplus :\Bbb{R}_+ \times \Bbb{R}_+\to \Bbb{R}_+ \qquad x\oplus y =x\cdot y\\ \odot :\Bbb{R}_+ \times \Bbb{R}_+\to \Bbb{R}_+ \hspace{0mm}\qquad a\odot x =x^a\hspace{3.5mm}$$

From this, I think we can construct - for example - the $2-$dimensional vector space with the following operations:

$$\oplus :\Bbb{R}_+^2 \times \Bbb{R}_+^2\to \Bbb{R}_+^2 \qquad (x_1,y_1)\oplus (x_2,y_2) =(x_1\cdot x_2,y_1 \cdot y_2)\\ \odot :\Bbb{R}_+ \times \Bbb{R}_+^2\to \Bbb{R}_+^2 \hspace{20mm} a\odot (x_1,y_1) =(x_1^a,y_1^a)\hspace{15mm}$$

  • How do we define (if possible) a inner product in this vector space? Just looking from the definition, it's not clear to me how to do it or if it's possible.

  • Could we use the classical inner product $(x_1,y_1)\cdot (x_2,y_2)=x_1 x_2 + y_1 y_2$? I think that as we're using $\Bbb{R}_+$, vectors would never be orthogonal? I'm not sure if this is an actual problem.

Red Banana
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1 Answers1

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This vector space is just a disguised form of $\mathbb{R}^2$, so it admits the usual inner product that $\mathbb{R}^2$ does but you have to disguise it appropriately. The isomorphism is given by taking the logarithm (and in the opposite direction by exponentiating) so the inner product is

$$(x_1, y_1) \cdot (x_2, y_2) = \log x_1 \log x_2 + \log y_1 \log y_2 \in \mathbb{R}.$$

For more on this the keyword is transport of structure. Also, you want to be a little careful about scalar multiplication: the scalar here still takes values in $\mathbb{R}$ as usual, not $\mathbb{R}_{+}$, unless you want to confuse yourself even more by applying an additional logarithm / exponential.

$x_1 y_1 + x_2 y_2$ isn't bilinear in this case; the whole point of this construction is that addition and scalar multiplication are different from normal!

In general, every real vector space admits an inner product. This is because, by the axiom of choice, every real vector space $V$ admits a basis $e_i, i \in I$ (where the index set $I$ may be infinite), and then we can define the inner product by linearly extending $\langle e_i, e_j \rangle = \delta_{ij}$ in the usual way, so that the $e_i$ form an orthonormal basis. However, this generally isn't a very useful or interesting construction if $V$ is infinite-dimensional, and usually we want to pick other inner products which are better suited to applications.

Qiaochu Yuan
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