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Is the following correct ?? And is there a proof for this ?

$\| c - d\|^2 = \| c\|^2 + \|d \|^2 - 2cd$

Another question: Why is the following true ?? I do not understand how it goes from LHS to RHS. (Notice the thing below actually comes from the proof of Jame-Stein Estimator)

$\left\| \frac{X}{\|X\|^2} \right\|^2 = \frac{1}{\|X\|^2}$

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What vector spaces are you talking about? "2cd" might not be well defined. For the other remember that $\frac 1 {\lVert X \rVert^2}$ is a real number. –  Stefan Dec 27 '12 at 1:38
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$$\left\|\frac{X}{\|X\|^2}\right\|^2 = \left\|\frac{X}{\|X\|^2}\right\|\cdot\left\|\frac{X}{\|X\|^2}\right\| = \left(\frac1{\|X\|^2}\|X\|\right)\left(\frac1{\|X\|^2}\|X\|\right) = \frac1{\|X\|}\cdot\frac1{\|X\|}=\frac1{\|X\|^2}$$ –  Brian M. Scott Dec 27 '12 at 1:39
    
As Stefan pointed out, not all things that have a norm can be multiplied. For the case of $\mathbb{R}^n$, this is essentially the law of cosines. –  Karolis Juodelė Dec 27 '12 at 1:53

1 Answer 1

To elaborate on the first question: $$\lVert c-d\rVert^2 = \langle c-d, c-d \rangle = \langle c,c \rangle - 2 \Re(\langle c,d\rangle )+ \langle d,d\rangle = \lVert c \rVert^2 + \lVert d \rVert ^2 -2cd$$ but only if $cd := \langle c,d \rangle$ and we are talking about a real vector space where the norm is induced by the inner product. So for $\mathbb R^n$ your equality holds.

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