# Visualization of length and orthogonality under non-standard inner product

In $\mathbb{R}^3$ under the standard inner product, the length of

$$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$

is 1. However, under the innner product

$\left&space;\langle&space;\overrightarrow{x},&space;\overrightarrow{y}&space;\right&space;\rangle&space;=&space;a&space;x_1&space;y_1&space;+&space;x_2&space;y_2&space;+&space;x_3&space;y_3$

where a>0, the length becomes

$\left&space;\|&space;e_1&space;\right&space;\|&space;=&space;\sqrt{\left&space;\langle&space;e_1,&space;e_1&space;\right&space;\rangle}&space;=&space;\sqrt{&space;a&space;\times&space;1&space;\times&space;1+0\times&space;0+0\times&space;0}&space;=&space;\sqrt{a}$

It is obvious to me that under the standard inner product the unit vector has length of 1; also I know the above calculation of the length under the non-standard inner product is following the definition of length. However, how would one visualize the length of the unit vector being $\sqrt{a}$ under the non-standard inner product defined above ?

Similarly, I have problem visualizing the fact that

$$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$

and

$$\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$$

are not orthogonal to each other under the non-standard inner product.

• For the first part, you could perhaps tihink of a tape measure along the $x$-axis marked in both metres and feet? For the second part, those two vectors are orthogonal under the particular non-standard metric you have given. Feb 10, 2021 at 22:17
• Thanks for your comment. Those two are indeed orthogonal - it was my mistake. I was meant to write some other metric. Also you mentioned metres and feet - it starts to make a little sense to me now. Feb 10, 2021 at 22:43

While vectors under norm generated by standard inner product can be visualized on $$n$$ dimensional sphere, the non-standard products leads to $$n$$ dimensional ellipsoids.
You can rewrite the inner product as $$x^TAy$$ where $$A$$ is positive definite matrix. For setting $$A = I$$ you have standard inner product. An expression $$x^TAy$$ is so-called quadratic form which can be visualised as an ellipsoid (under assumption that $$A$$ is positive definite which is also necessary to fulfil axioms of inner product).
In your example (I switched to 2D only but this can be generalized for $$n$$ dimensions) vector $$\begin{pmatrix} 1 & 0 \end {pmatrix}$$ is unit vector which lays on x axis, vector $$\begin{pmatrix} 0 & 1 \end {pmatrix}$$ layes on y axis and ends of these vectors lays on unit sphere with centre in point (0,0). Under your non-standard inner product, vector $$\begin{pmatrix} 1 & 0 \end {pmatrix}$$ lays again on x axis and also on major axis of ellipse, vector $$\begin{pmatrix} 0 & 1 \end {pmatrix}$$ lays again on y axis and also on minor axis of ellipse. The ellipse is centered in point (0,0). A length of the ellipse major axis is $$\sqrt{2}$$ and minor axis has length 1.