I know that we can define an ellipsoid by its center and a set of orthongonal vecotrs. Can we also uniquely define an ellipsoid in the following manner:
Let $B$ be some orthogonal basis of $\mathbb R^n$. Let $c$ be the ellipsoid's center. Assume now that the lengths of the vectors in $B$ satisfy that $\forall v\in B : c+v\in \partial E$, (where $\partial E$ is the ellipsoid's boundary).
Does the pair $(c,B)$ define the ellipsoid uniquely?
If so, what is the matrix $A$ that satisfies $E=\{x\in\mathbb R^n| (x-c)^T A (x-c)\leq 1\}$ ?
Edit
I was told in the answers that it doesn't uniqely define the ellipsoid, so I change my question a bit:
We now have triple $(c,\{v_1,\dots,v_n\},\{N_1,\dots,N_n\})$, such that $c$ is the center, and $B$ is now replaced with $\{v_1,\dots,v_n\}$, and $\{N_1,\dots,N_n\}$ is a set of normals to the ellipsoid at the points $\{v_1,\dots,v_n\}$. That is, $\forall i\ : v_i\perp N_i$. And additionally, $\forall i: c + v_i\in\partial E$ as before.
Does this triple uniquely define the ellipsoid?
Thanks in advnance!
Let $E$ be the set of points in $\mathbb R^2$ satisfying $x^2 - xy + y^2 = 1$. Then the center of $E$ is $c = (0,0)$ and $B = \{(1,0), (0,1)\}$ is an orthogonal basis of $\mathbb R^2$ satisfying the condition that $\forall v\in B : c+v\in \partial E$.
But $B$ also satisfies the corresponding condition for the set of points satisfying $x^2 + xy + y^2 = 1$, for the unit circle, and for numerous other elliptical regions.
You can easily extend this to higher dimensions. For example, start with an ellipsoid in $\mathbb R^n$ with one axis of length $a$ and $n-1$ axes of length $b$, where $a > b$ the axis of length $a$; take an orthogonal basis of $\mathbb R^n$ in which no axis is parallel to the ellipsoid's axis of length $a$; and multiply each vector of the orthogonal basis in order to construct a set $B$ satisfying $\forall v\in B : c+v\in \partial E$. Now you can find a distinct ellipsoid $E' \neq E$ such that $\forall v\in B : c+v\in \partial E$, simply by reflecting $E$ through one of the hyperplanes through $c$ parallel to the span of $n-1$ members of $B$ or by using $B$ as the set of principal axes of the new ellipsoid.
So in the sense that I understand the question, that you are given an arbitrary ellipsoid $E$ and choose the orthogonal basis $B$ without imposing any additional conditions to guarantee that all its vectors are parallel to principal axes of $E$, the basis $B$ does not uniquely specify the original ellipsoid $E$.
To put it in terms of the matrix $A$, you can always find a diagonal matrix that will satisfy $\forall v\in B : (v-c)^T A (v-c) = 1$, but there will also be a matrix $A'$ with non-zero terms off the diagonal such that $\{x\in\mathbb R^n| (x-c)^T A' (v-c)\leq 1\}$ is an ellipsoid, and then $\forall v\in B : (v-c)^T A' (v-c) = 1$.
If the basis $B$ is required somehow to have every vector parallel to a principal axis of $E$, then I believe it specifies $E$ uniquely.
If in addition to $c$ and the basis $B = \{v_1,\dots,v_n\}$, you find the normal $N_i$ to $\partial E$ at each point $c + v_i$, then it seems "obvious" that the tuple of vectors $(c,v_1,\dots,v_n,N_1,\dots,N_n)$ uniquely determines the ellipsoid $E$ from which it was obtained, in the sense that it is "obvious" that there is only one linear transformation $M$ that takes the unit sphere to $E - c$ and that any other linear transformation of the unit sphere will map the neighborhood of at least one of the $M^{-1}v_i$ to a different surface. I think this is also true when the correspondence between the $v_i$ and the $N_i$ is not explicit, such as if we write $(c,\{v_1,\dots,v_n\},\{N_1,\dots,N_n\})$, because it seems "obvious" there is only one way to map the $N_i$ to the $v_i$ such that the $N_i$ are normal to an ellipsoid through $\{v_1,\dots,v_n\}$. In fact I think the ellipsoid is then a bit "overspecified", since there are many sets $\{N_1,\dots,N_n\})$ that could not be the normals to any ellipsoid at the points $c + v_i$, and I think at least one of the $N_i$ can be omitted without losing uniqueness. How to prove these "obvious" facts escapes me at the moment, however.