In this example,
It is saying that the 2 vectors
$ \frac{(i,1)}{ \sqrt 2} , \frac {(-i,1)}{ \sqrt 2}$
are an orthonormal basis, I wanted to verify this, so I checked the norm of each vector, the norms turned out to be $1$
But when I perform the inner product of the two vectors
$\langle \frac{(i,1)}{ \sqrt 2},\frac {(-i,1)}{ \sqrt 2}\rangle = 1$
It does not equal zero, so they should not be orthonormal, is this a typo in the book or am I missing something?
On
Inner products on complex vector spaces are customarily not entirely as straight-forward as they are for real spaces.
One of the properties we would like an inner product to have is that the inner product of a vector with itself should be a positive real number. This is clearly not the case for, say, the vector $(i,0)$ if you use the same formula as the one you're used to from real linear algebra.
The conventional inner product on complex spaces conjugates the first factor before computing the inner product the "real" way. For instance, $$ \langle(i,0),(i,0)\rangle=(\bar i\cdot i)+(\bar 0\cdot0)=1 $$ Applying this to your inner product gives a result of $0$.
We have that $$\left\langle[i,1]^t, [-i, 1]^t\right\rangle=\bar{(i)}(-i)+\bar{(1)}(1)=(-i)^2+1=0$$ The same applies when the inner product is linear in it's first argument instead of the second.