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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 $\|A-B\|^2?$

user29999
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    Whenever the norm is induced by an inner product $\langle ,\cdot,,,\cdot,\rangle$; we can write $$\lVert x-y\rVert^2 =\lVert x\rVert^2-2\langle x,y\rangle+\lVert y\rVert^2$$ Note your two equations are particular cases of this where $\langle x,y\rangle =x_1y_1+\cdots+x_ny_n$. In particular $\langle x,y\rangle=xy$ is an inner product in $\Bbb R$. – Pedro Aug 01 '13 at 18:34
  • Is $|A|$ induced by an inner product? What is $\langle A,B\rangle?$ – user29999 Aug 01 '13 at 19:03

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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.

user7530
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  • Very good your observation, thank you. But you do not answer my ask yet. My ask is motivated by the fact that in my calculations appear something like $|A-B|$ and an estimate for this. – user29999 Aug 01 '13 at 19:21
  • He did, if your norm comes from an inner product, the formula holds, if not, then what is $x\cdot y$ supposed to be? – Tomas Aug 01 '13 at 19:25
  • An general estimate would be $|,||A||-||B||;|\leq ||A-B||\leq ||A||+||B||$. – Tomas Aug 01 '13 at 19:32