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i would need an explanation with an example on why is the inner product defined as it is and not like a dot product:

\begin{align} \text{inner pr.:}& & &\boxed{a\cdot b = a_1\overline{b_1} + a_2\overline{b_2}+...a_n\overline{b_n}}\\ \text{dot pr.:}& & &\boxed{a\cdot b = a_1b_1 + a_2b_2+...a_nb_n} \end{align}

Here it says:

For vectors with complex entries, using the given definition of the dot product would lead to quite different properties. For instance the dot product of a vector with itself would be an arbitrary complex number, and could be zero without the vector being the zero vector.

Can anyone provide me an example on this and show me how the definition for inner product above solves this?

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1 Answer 1

up vote 4 down vote accepted

Hint: With $u = (1+i, 0)$, what is the dot product $u\cdot u$?

Hint: With $v = (1, i)$, what is the dot product $v \cdot v$?

How do we classify all 2-d vectors such that $v \cdot v = 0$?

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Thank you. Is there any good case for 3-D vectors describing this? I need 2 different 3-D vectors and a proof of this on them. –  71GA May 13 '13 at 1:36
    
I get $u \cdot u = 2i$ if i calculate it like a dot product and i get $u \cdot u = 2$ if i calculate it as a inner product. Then for $v$ i got $v\cdot v = 0$ if i calculated it as a dot product and $v\cdot v = 3$ if i calculated it as an inner product. Is this good justification for using definition for inner product as it is? –  71GA May 13 '13 at 7:33
    
@71GA It's not that hard to create 3-D vectors once you understand this example. –  Calvin Lin May 13 '13 at 17:28
    
I am trying to think that an inner product can only be aplied to vectors so first i need a complex vector. And i try to think of a complex vector as a column matrix with complex numbers (at least one has to be ) like this one: $$\vec{v} = \begin{pmatrix}1+3i\\2-i\\ 3\end{pmatrix}$$ and here complex numbers are $v_1=1+3i,\, v_2=2-i,\,v_3 = 3$. So now i can use theese to calculate inner product $v\cdot \overline{v} = v_1\overline{v_1}+v_2\overline{v_2}+v_3\overline{v_3}$ which is always real (and this is what we wanted it to be so now i understand why this definition is OK). –  71GA May 13 '13 at 19:37
    
I forgot to say that inner product $v\cdot \overline{d}$ isn't demanded to be $\mathbb{C}$, so there isn't any problem here anymore. Is my thinking mathematically correct? –  71GA May 13 '13 at 19:38

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