Is this true for the inner products ? :
$(\vec a + \vec b)\cdot(\vec c + \vec d) = \vec a\cdot\vec c + \vec a\cdot\vec d + \vec b\cdot\vec c + \vec b\cdot\vec d$.

  • $\begingroup$ It distributes like: $(\vec{a}+\vec{b})\cdot\vec{c}=\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}$. $\endgroup$ – MattAllegro Apr 18 '14 at 19:27
  • $\begingroup$ This should be true for any inner product - not just the standard one defined for vectors but also for example those defined for continuous functions. It should follow from the properties of the inner product. $\endgroup$ – AnlamK Oct 1 '17 at 20:12

It is indeed true. It should be easy to confirm this by writing out the vectors explicitly in component form and evaluating both sides of the equality.

Remember that if $\mathbf{x} = \langle x_1, x_2, ..., x_n \rangle$ and $\mathbf{y} = \langle y_1, y_2, ..., y_n \rangle$, then:

$$\mathbf{x} \cdot \mathbf{y} = \sum_{k=1}^n x_ky_k$$

  • $\begingroup$ Ok thank you ! Do you know if all the "classic" distributive properties applicable to scalar products ? $\endgroup$ – Trevör Apr 18 '14 at 19:24
  • 1
    $\begingroup$ Indeed. The associative, distributive, and commutative law all hold. Even the product rule holds when taking derivatives! $\endgroup$ – Kaj Hansen Apr 18 '14 at 19:30

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