What is the relation between these two definitions of an ellipsoid There are two definitions of an ellipsoid in Boyd's book (Convex Optimization)


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*$E = \{ x | (x-x_c)^T P (x-x_c) \leq 1 \}$


In the above, P is a positive semi definite matrix.


*$ E=\{ x_c+Au  |\; ||u|| \leq 1  \}$ 


In the second definition, $A$ is non-singular.
I have certain questions about the above definitions.


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*Why is positive definite required for the first case? what if $P$ is not positive semi definite?

*I know that non-singularity is required for $A$, otherwise ellipsoid dimensionality would reduce. I want to know that what is the relation between these two definitions. Is there any relation between $A$ and $P$?

 A: I'll consider only vector spaces over $\mathbb R^n$ if that's OK.
Since $A$ is non-singular, $A^{-1}$ exists and is non-singular.
So $x = Au + x_c$ is equivalent to $u = A^{-1}(x - x_c)$,
and we can rewrite the second definition as follows:
$$ E = \{ x \mid  \|A^{-1}(x - x_c)\| \leq 1  \}.$$
But 
$\|A^{-1}(x - x_c)\| = (x - x_c)^T \left(A^{-1}\right)^T A^{-1}(x - x_c)$,
and since $A^{-1}$ is non-singular,
$\left(A^{-1}\right)^T A^{-1}$ is positive definite.
In fact, if we let $P = \left(A^{-1}\right)^T A^{-1}$,
then $P$ is the positive definite matrix
you need for the first definition: 
$(x - x_c)^T P (x - x_c) = \|u\|$ where $u = A^{-1}(x - x_c)$,
hence the set of $x$ such that 
$(x - x_c)^T P (x - x_c) \leq 1$ is just the set of 
$x = Au + x_c$ such that $\|u\| = 1$.
A: The condition that $P$ be positive definite, is equivalent to saying that all of it's eigen values are positive. This means, that if you take the eigendecomposition, $P= U^{-1}DU$, where $D$ is a diagonal matrix and $U$ unitary, $D$ will have only positive values.
If you think of the unitary $U$ as a combination of rotation and stretching, you see that your first equation reduces to taking an axis aligned ellipsoid $\sum a_i x_i^2 \le 1,\  a_i > 0$, and then rotating and stretching it.
If $P$ is not positive definite, then some of the values of $D$ will be negative, leading to a hyperbola or some other shape (e.g. $x^2-y^2 =1$).
