Linear mappings between two Sets For $n\ge 2$ consider the following two subsets of the inner product space $\mathbb{C}^n$ with usual Euclidean inner product denoted by $\langle\,,\mkern2mu\rangle$
$$B=\{x\in\mathbb{C}^n:\|x\|<1\}$$
and,
$$E=\{x\in\mathbb{C}^n:\sum_{i=1}^{n}|x_i|^{2p_i}<1\}$$
where each $p_i$ is a real number greater than or equal to 1 and not all $p_i$ are 1 and $x_i$ is the $i^\text{th}$ coordinate of $x$.
I am trying to show that there is no invertible linear operator $T$ on $\mathbb{C}^n$ such that $T(B)=E$.
This is what I have done so far :
Let $T^*$ be the hermitian adjoint of $T$ (ie $\langle Tx,y\rangle =\langle x,T^*y\rangle$ for all $x, y\in\mathbb{C}^n$) and let
$$(TT^*)^{-1}=\sum_{i=1}^{k}r_iP_i$$
be the spectral decomposition of the positive definite operator $(TT^*)^{-1}$. A simple argument shows that
$$T(B)=\{x\in\mathbb{C}^n:\sum_{i=1}^{k}r_i\|P_ix\|^2<1\}.$$
$T$ must be a homeomorphism and hence it must preserve boundaries. Thus it is sufficient to show that there is some $x\in\mathbb{C}^n$ which satisfies only one the following equations
\begin{equation}
\sum_{i=1}^{k}r_i\|P_ix\|^2=1
\end{equation}
$$\sum_{i=1}^{n}|x_i|^{2p_i}=1$$
Another idea that I have in mind is to somehow deduce that each $r_i=1$. Spectral theorem then would imply that
$$\sum_{i=1}^{k}r_i\|P_ix\|^2=\sum_{i=1}^{n}|x_i|^2$$
 A: A convenient framework for your problem is Convex Geometry theory. As Andrew D. Hwang pointed out, a linear image of a Euclidean ball is -- by definition -- an ellipsoid. If you like, think about an ellipsoid in $\mathbb{C}^n$ (or $\mathbb{R}^n$) as a unit-ball of an inner-product space.
In your specific problem, the set $E$ is convex and centrally symmetric with respect to the origin. This means that $v\in E$ if and only if $-v\in E$. Its closure is compact. Therefore, it is an interior of a unit-ball of some normed space, (whose norm can be written in terms of the Minkowski gauge of the closure of $E$, but we do not need it now).
If there were a one-to-one linear map $T$ such that $TB=E$, then by continuity of linear maps, $T$ would carry the closure of $B$ onto the closure of $E$, which - as A.D Hwang pointed out -- would then force $E$ to be an ellipsoid. What we know about ellipsoids is that every central section of them is an ellipsoid of the appropriate dimension. So if we pick an index $1\leq i\leq n$ for which $p_i>1$, and consider the intersection of $E$ with the complex line spanned by the unit vector $e_i$, ($1$ at the $i$'th spot, zeros at all the rest), we will have obtained an ellipse in a complex plane. On the other hand, the intersection of $E$ with the span of $e_i$ is given by
$$\{x\in\mathbb{C}: |x|^{2p_i}<1\}$$
In other words, we get the interior of the contour in the plane whose equation in Cartesian (real) coordinates $(x,y)$ is
$$(x^2+y^2)^{p_i}=1$$
That this is not ellipse should be clear, because $p_i>1$. This proves that there can be no one-to-one linear map carrying $B$ onto $E$. There is of course a non-linear homemorphism that does the job, because every two convex full-dimensional sets containing the origin are homeomorphic.
A different approach to the problem is to consider the group of linear symmetries of $B$ and $E$ respectively. I am pretty sure that the group of linear symmetries of $E$, i.e. the set of linear maps that leave $E$ invariant, consists of permutations of the coordinates and multiplication of the coordinates by complex numbers of absolute value one. However, I have no proof. On the other hand, the set of symmetries of $TB$ is obtained by conjugating the entire unitary group of orthogonal transformations by $T$, which surely is much larger than permutations combined with multiplications by $e^{i\theta}$'s. In any case, the argument above is probably easier.
