Is this complex vector orthogonal to itself?

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SUMMARY

The basis vector (i,0,1) in the space V=Span((i,0,1)) with a standard inner product over ℂ³ is not orthogonal to itself. The inner product calculation <(i,0,1),(i,0,1)> yields zero, indicating a misunderstanding of the complex inner product properties. In complex vector spaces, the inner product must be anti-linear in one argument, which clarifies that the inner product is defined as ⟨x, y⟩ = ∑k xk* yk.

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Steve Turchin
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Is the basis vector ##(i,0,1)## in the space ##V=##Span##((i,0,1))## with a standard inner product,over ##\mathbb{C}^3##
orthogonal to itself?
##<(i,0,1),(i,0,1)> = i \cdot i + 0 \cdot 0 + 1 \cdot 1 = -1 + 1 = 0 ##
The inner product (namely dot product) of this vector with itself is equal to zero.
What is going on here?
 
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No, it is not orthogonal to itself. The inner product on a complex vector space must be anti-linear in one of the arguments (which one depends on whether you use physics or maths notation). In other words, the complex inner product is given by
$$
\langle x, y \rangle = \sum_k x_k^* y_k.
$$
 
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