Inner Product of 0 vector, & Complex numbers

[DeepSeeded]

Hello,

Can someone help me understand why the Inner Product of a Null vector with itself can be non zero if complex numbers are involved?

And why using the complex conjugate resolved this?

I may have understood this wrong. It could be that an Inner Product of any non-Null vector with itself can be zero if complex numbers are involved. Either way does someone have an example?

 

[dx]

It cannot. The inner product of the null vector with itself is zero, and the inner product of a non-null vector cannot be zero. Both these facts follow from the following axiom for inner products:

$$\langle v|v \rangle \geq 0$$, and $$\langle v|v \rangle = 0$$ if and only if $$v = |0\rangle$$​

 

[HallsofIvy]

The inner product on a vector space over the complex numbers must satisfy
$$<u, v>= \overline{<v, u>}$$ and, in particular, $<u, av>= \overline{a}<u, v>$ for any scalar a.

For C², ordered pairs of complex numbers, the inner product $<(a+ bi),(a+ bi)> = (a+ bi)\overline(a+ bi)= (a+bi)(a-bi)= a^2+ b^2$ which is 0 only if a= b= 0.

If you mistakenly use the inner product for real numbers, that is, without the "complex conjugation" you could have $<a+ bi, a+ bi>= a^- b^2$ which can be 0. But, as I said, that is a mistake.

 

[DeepSeeded]

The inner product on a vector space over the complex numbers must satisfy
$$<u, v>= \overline{<v, u>}$$ and, in particular, $<u, av>= \overline{a}<u, v>$ for any scalar a.

For C², ordered pairs of complex numbers, the inner product $<(a+ bi),(a+ bi)> = (a+ bi)\overline(a+ bi)= (a+bi)(a-bi)= a^2+ b^2$ which is 0 only if a= b= 0.

If you mistakenly use the inner product for real numbers, that is, without the "complex conjugation" you could have $<a+ bi, a+ bi>= a^- b^2$ which can be 0. But, as I said, that is a mistake.

Ahha! Thank you!