Why do transformations of ellipses by matrix transformations remain ellipses. How do I show that given an ellipse which is transformed by a matrix it will remain an ellipse. (unless the matrix is not invert-able). You could consider a line segment or point as a special case of an ellipse. 
 A: Mostly because of this: say $H$ is a square symmetric, and positive definite, matrix. Let your ellipse, or ellipsoid be
$$ x^T H x = C  $$ for column vector $x$ and positive real constant $C.$ If it is also true that
$$  H = P^T G P  $$
with symmetric $G,$ then  $ x^T H x = C  $ becomes $x^T P^T G Px = C.$ Which means that when $x$ satisfies the one with $H,$ then $Px$ satisfies the one with $G$
A: For the purposes of this solution, a set of the form
$$
\mathcal{E}=\mathcal{E}(z,D) = \{ x \in \mathbb{R}^n \mid (x-z)^\top D (x-z) =1\}, 
$$
in which $D$ is a symmetric, positive semidefinite matrix, is called an ellipse centered at $z$. 

Proposition: If $A \in M_n(\mathbb{R})$ is an invertible matrix and $A(\mathcal{E}) := \{ y\in \mathbb{R}^n \mid y=Ax,~x\in \mathcal{E}\}$, then $A(\mathcal{E}) = \mathcal{E}(Az,A^{-\top}DA^{-1})$, in which $A^{-\top} := (A^\top)^{-1} = (A^{-1})^\top$.

Proof. If $y\in A(\mathcal{E})$, then $y=Ax$, $x\in \mathcal{E}$. Notice that
\begin{align*}
(y-Az)^\top A^{-\top}DA^{-1}(y-Az) 
&= ((x-z)^\top A^\top)A^{-\top}DA^{-1}(A(x-z)) \\
&= (x-z)^\top D(x-z) = 1.
\end{align*}
The result follows after showing that the matrix $A^{-\top}DA^{-1}$ is symmetric and positive semidefinite (simple exercise).  
