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One of the definitions of an ellipsoid in Boyd & Vandenberghe's Convex Optimization is

$$E = \{ x_c + A u : \| u \|_2 \leq 1\}$$

where $A$ is square and non-singular. It is also stated that if $A$ is singular, then we get a degenerate ellipsoid.

Can we generalize this set, that is, does the set $E$ still remain an ellipsoid if we take $A$ to be non-square? Or is there a special name for such sets?

Rodrigo de Azevedo
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Cherryblossoms
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    Yes, it is still an ellipsoid. The definition basically says that you map a unit ball to the range of $A$, which is by def. an ellipsoid. It is always a degenerate ellipsoid in the non-square cases. – V.S.e.H. Mar 04 '21 at 10:53
  • @V.S.e.H.: if $A$ is wider than it is tall, the result doesn't need to be degenerate. – Troposphere Apr 09 '21 at 17:15
  • @Troposphere You might want to rethink. – V.S.e.H. Apr 09 '21 at 19:37
  • @V.S.e.H.: Why? If, for example, $x_c=\begin{pmatrix}0\0\0\end{pmatrix}$ and $A = \begin{pmatrix}1&0&0&0\0&1&0&0\0&0&1&0\end{pmatrix}$, then $E$ becomes the usual closed unit ball in $\mathbb R^3$, which seems to be a perfectly good non-degenerate ellipsoid by the original defintion too. – Troposphere Apr 09 '21 at 22:09
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    @Troposphere Your $A$ represents a map from $\mathbb{R}^4$ onto $\mathbb{R}^3$, not from $\mathbb{R}^3$... – V.S.e.H. Apr 10 '21 at 19:25
  • @V.S.e.H.: Indeed it does. It is necessary for speaking about any non-square $A$ in the first place that $u$ is going to live in a space of a different dimension than the space $E$ is a subset of. – Troposphere Apr 10 '21 at 19:29
  • @Troposphere You're right, if we agree on the definition that a degenerate ellipsoid has an empty interior (which it doesn't in your case). – V.S.e.H. Apr 10 '21 at 20:46

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