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Compute inner product $\langle (1+i,4) , (2-3i, 4+5i) \rangle$ in $\mathbb{C}^2$

  • My Answer:

$= (1+i)(2-3i) + 4(4+5i)$

  • Book's answer:

$=(1+i)(2+3i)+4(4-5i)$

Why does the book's method conjugate the terms in the second argument of the inner product?

I didn't know we conjugated when directly computing the standard inner product over a complex field. I thought we only conjugate when pulling out a scalar from the second argument or when we swap the order of arguments.

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There are books that define: $$\langle z,w\rangle_1 = \sum_{j=1}^nz_j\overline{w_j},$$ while others define $$\langle z,w \rangle_2 = \sum_{j=1}^n\overline{z_j}w_j,$$so you have to pay attention to that. If we don't conjugate one of the terms, we won't have the property $\langle z,w\rangle = \overline{\langle w,z\rangle}$ we want. Both are called the standard inner products in $\Bbb C^n$. I numbered them so we can compare: $$\langle (1+i,4),(2-3i,4+5i)\rangle_1 = (1+i)(2+3i)+4(4-5i) \\ \langle (1+i,4),(2-3i,4+5i)\rangle_2 = (1-i)(2-3i)+4(4+5i)$$

You forgot to take conjugates there. Your book is using $\langle\cdot,\cdot\rangle_1$.

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  • $\begingroup$ How are they allowed to use different definitions for a standard inner product when they produce different answers? Also, how am I supposed to tell which one is used - do I look for the subscript? $\endgroup$ – David South Jun 4 '15 at 2:14
  • $\begingroup$ Nevermind, I get it! Thank you! $\endgroup$ – David South Jun 4 '15 at 2:15
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    $\begingroup$ They won't use any subcript. I put them here for the sake of the answer. The point is that we can have a lot of possible inner products for a single space. In some point of the book, they'll have to explictly give you the expression, at least once. $\endgroup$ – Ivo Terek Jun 4 '15 at 2:15
  • $\begingroup$ They still produce different answers. How is this acceptable? $\endgroup$ – David South Jun 4 '15 at 2:17
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    $\begingroup$ They produce different answers because they are different inner products. .. The moral of the story is: pick one of them and stick to it. $\endgroup$ – Ivo Terek Jun 4 '15 at 2:18
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You have $\langle a, b\rangle = \sum_i a_i\overline{b_i}$ (or $\langle a, b\rangle = \sum_i \overline{a_i}b_i$ depending on the day of the week) so that $\langle a, a\rangle = \|a\|^2$.

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    $\begingroup$ Protip: using \langle and \rangle instead of < and > produces better symbols. The same goes for \overline{} instead of \bar{} :) $\endgroup$ – Ivo Terek Jun 4 '15 at 2:26
  • $\begingroup$ Thanks. Upvoted. Now all I have to do is remember these. With $\LaTeX$, I'm happy if the results are even close to what I want. $\endgroup$ – marty cohen Jun 4 '15 at 2:42

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