Add $l^2$ norms with equality Let $x \in \mathbb{R}^n$ be some vector and define matrices (not necessarily square) $A \in \mathbb{R}^{n_A \times n}$, $B \in \mathbb{R}^{n_B \times n}$.
My goal is to express the sum of the $l^2$ norms of the vectors $Ax$ and $Bx$ in the following form:
\begin{equation}
   \lVert Ax \rVert_2 + \lVert Bx \rVert_2 = \sqrt{x^T C^T C x} = \lVert Cx \rVert_2
\end{equation}
where $C \in \mathbb{R}^{n_C \times n}$ is a function of $A$ and $B$ only.
n=1
In $\mathbb{R}^1$, this is trivial. For $x \in \mathbb{R}$,
\begin{equation}
    \lVert \vec{a} x \rVert_2 = \lVert \vec{a} \rVert_2 \lvert x \rvert
\end{equation}
Hence, for any $a \in \mathbb{R}^{n_A}$, $b \in \mathbb{R}^{n_B}$:
\begin{equation}
   \lVert \vec{a} x \rVert_2 + \lVert \vec{b} x \rVert_2
   = (\lVert \vec{a} \rVert_2 + \lVert \vec{b} \rVert_2) \lvert x \rvert
   = \sqrt{C^2x^2}
\end{equation}
where $C = \sqrt{\lVert a \rVert_2 + \lVert b \rVert_2} \in \mathbb{R}$.
Is this possible for $n > 1$? Under what conditions can this be done?
 A: It is true if and only if $P = A^T A$ is a constant multiple of $Q = B^TB$.
The implication from right to left is straightforward.  Let's do the other way.
Case 1: One of $P$ and $Q$ is positive definite.
It is well known that two quadratic forms, at least one of which is positive definite, can be simultaneously diagonalized.  That is, there is a basis of vectors $x_1,x_2,\dots,x_n$ such that $x_i^T P x_j = x_i^T Q x_j = 0$ for $i \ne j$.
If $P$ is not a constant multiple of $Q$, then there exist two vectors $u$ and $v$ such that $u^T P u = v^T P v = 1$ and $u^T Q u = \lambda \ne v^T Q v = \mu$ and $u^T P v = u^T Q v = 0$.
Let $ {\|x\|}_* = \sqrt{x^T P x} + \sqrt{x^T Q x} $.
We want to show that there doesn't exists a matrix $C$ such that if ${\|x\|}_*^2 = x^T C x$.  For if such a matrix did exist, then we would have the polarization identity:
$ {\|u+v\|}_*^2 + {\|u-v\|}_*^2 = 2{\|u\|}_*^2 + 2{\|v\|}_*^2 $.
Do the calculations, and you will see that this identity holds if and only if $\lambda = \mu$.
Case 2: Both $P$ and $Q$ are not positive definite.  Since they are positive semi-definite, this means that they both have non-trivial kernels.  If one kernel is contained in the other, then simply restrict the quadratic forms to the orthogonal complement of the smaller kernel, and repeat the above argument.  If one kernel is not contained in the other, then there exists $u$ and $v$ such that $u^T P u = 1$, $u^T Q u = 0$, $v^T P v = 0$, $v^T Q v = 1$.  Then the polarization identity can again be seen to fail.
A: If $B=kUAV$ for some scalar $k$ and orthogonal matrices $U$ and $V$, then
$$\|Ax\| + \|Bx\| = (1+|k|) \|Ax\| = \sqrt{x^\top C^\top C x}$$
for $C=\sqrt{1+|k|}A$.
I am not sure if this is the most general condition on $A$ and $B$ so that your equality holds.
