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I am trying to figure out the reason for this line of deduction ( It is a proof for Householder's transformation on some vector) $$\|\mathbf{v}-\mathbf{u}\|^2= \langle \mathbf{v},\mathbf{v}\rangle- \langle \mathbf{v},\mathbf{u} \rangle-\langle \mathbf{u},\mathbf{v} \rangle+\langle \mathbf{u},\mathbf{u} \rangle = -2\langle\mathbf{v}-\mathbf{u},\mathbf{u} \rangle $$

How did they deduce the first and the second equality? Do we use a geometrical aid to help us think about this?

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    $\begingroup$ +1 for mentioning the context; without it, your second equality would be very confusing. $\endgroup$ – Ian Jul 13 '15 at 9:45
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    $\begingroup$ Because she's just not cool, man. $\endgroup$ – RBarryYoung Jul 13 '15 at 16:44
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The first equality: By definition $\|\mathbf v-\mathbf u\|^2=\langle \mathbf v-\mathbf u,\mathbf v-\mathbf u\rangle$, and then you just expand that using linearity of the inner product to the left and the right.

The second is not true in general, except if you have an assumption that $\langle\mathbf v,\mathbf v\rangle=\langle\mathbf u,\mathbf u\rangle$.

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    $\begingroup$ In the context mentioned in the question, that assumption is probably actually that $\langle v,v \rangle = \langle u,u \rangle = 1$. $\endgroup$ – Ian Jul 13 '15 at 9:44
  • $\begingroup$ Thank you. Yes, the assumption that $\langle \mathbf{v},\mathbf {v}\rangle =\langle \mathbf{u},\mathbf {u}\rangle $ is true. "Linearity" was the keyword that explain both deductions to me. A more complete way to think about this: Consider $ \langle \mathbf{v}-\mathbf {u}, \mathbf{a}-\mathbf {b} \rangle $ , observe element product from the first row, ie $(v_{i} - u_{i})\times (a_{i} - b_{i}) = v_{i}a{i}- v_{i}b_{i}- u_{i}a_{i} +u_{i} b_{i}$ ... and then you can deduce that in dot product form $\endgroup$ – tintinthong Jul 14 '15 at 1:24
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    $\begingroup$ @tintinthong: Yes, but splitting into components applies only for the standard dot product. If you have a different inner product, you need to work directly with the axioms for inner products, such as $\langle a,b+c\rangle = \langle a,b\rangle +\langle a,c\rangle$ and so forth. $\endgroup$ – Henning Makholm Jul 14 '15 at 9:47

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