Inner Product of Complex Vectors?

[kq6up]

I was reading in my textbook that the scalar product of two complex vectors is also complex (I assuming this is true in general, but not in every case). However for the general definition (the inner product), each element of one of the vectors needs to be its complex conjugate. I learned this in Mary Boas' methods. If this is the case, shouldn't all the imaginary component get zapped, and leave a scalar in $\mathbb{R}$.

Chris

 

[mathman]

If A and B are different complex vectors, then (A,B) imaginary part will survive. Only for (A,A) or something similar (such as (A,B) where B is a real multiple of A) will the imaginary part = 0.

 

[kq6up]

Ok, so in general the inner product is a $\mathbb{C}$.

Thanks,
Chris

 

[jbunniii]

Consider the one-dimensional case, where the complex vectors are simply complex numbers $u$ and $v$, and the inner product is $\langle u, v \rangle = u\overline{v}$. This won't be real unless $u$ and $v$ have the same (or opposite) angles.

 

[Fredrik]

I was reading in my textbook that the scalar product of two complex vectors is also complex (I assuming this is true in general, but not in every case). However for the general definition (the inner product), each element of one of the vectors needs to be its complex conjugate. I learned this in Mary Boas' methods. If this is the case, shouldn't all the imaginary component get zapped, and leave a scalar in $\mathbb{R}$.

You can define an inner product on $\mathbb C^n$ by $\langle x,y\rangle =x^\dagger y$, where $x^\dagger$ is the complex conjugate of the transpose of x. There's no reason for the imaginary part of $\langle x,y\rangle$ to be zero (for arbitrary x,y), and it's not hard to think of a counterexample, but we have
$$\langle x,x\rangle =x^\dagger x=\sum_{k=1}^n x_k^*x_k=\sum_{k=1}^n|x_k|^2,$$ and the imaginary part of this is clearly zero.