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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
    Commented Jul 13, 2015 at 9:45
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    $\begingroup$ Because she's just not cool, man. $\endgroup$
    – RBarryYoung
    Commented Jul 13, 2015 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
    Commented Jul 13, 2015 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
    Commented Jul 14, 2015 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. $\endgroup$
    – hmakholm left over Monica
    Commented Jul 14, 2015 at 9:47

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