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.
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.