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In under the standard inner product, the length of

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

is 1. However, under the innner product

where a>0, the length becomes

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 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.

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    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Feb 10, 2021 at 22:17
  • $\begingroup$ 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. $\endgroup$
    – sofname
    Feb 10, 2021 at 22:43

1 Answer 1

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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.

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    $\begingroup$ Thanks. Looking at the inner product in quadratic form makes more sense: the matrix is effectively stretching in the first dimension. $\endgroup$
    – sofname
    Feb 12, 2021 at 22:05
  • $\begingroup$ @sofname: You are welcome. If the answer is OK for you, could you please accept it? $\endgroup$ Feb 20, 2021 at 7:43

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