$\|A-B\|^2 = ?$ We know that if $x,y \in \mathbb{R}$
\begin{equation}
(x-y)^2 = x^2 -2xy + y^2
\end{equation}
If $x,y$ are vectors in $\mathbb{R}^n$ we have
\begin{equation}
|x-y|^2=|x|^2 - 2 \ x \cdot y +|y|^2.
\end{equation}
where $x\cdot y$ is the usual scalar product. We know that there are several ways to define $\|A\|$ when $A$ is a matrix, for example
\begin{equation}
\|A\| = \sup  \{ |Ax|: |x|=1\}
\end{equation}
Is there a similar formula as above? This is a formula to $\|A-B\|^2?$

If the norm is induced by some inner product $\langle\cdot, \cdot\rangle$, it must obey the parallelogram law:
\begin{align*}
\|x+y\|^2 + \|x-y\|^2 &= \|x\|^2 + 2\langle x, y\rangle + \|y\|^2 + \|x\|^2 - 2\langle x, y\rangle + \|y\|^2\\
&= 2\|x\|^2 + 2\|y\|^2
\end{align*}
A first indication that something is wrong is that the two matrices $\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$ and $\left[\begin{array}{cc}1 & 0\\0 & 1\end{array}\right]$ have the same norm. Indeed,
$$\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]+\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2+\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]-\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2 = 2,$$
$$2\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]\right\|^2 +2\left\|\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2 = 4,$$
violating the law.
