What is the orthonormal basis of $a = \left(-\frac{3+i}{2}, 1\right)$ and $b = \left(\frac{3+i}{5}, 1\right)$?
What I did so far:
Let $A = 2a$ (for simplicity). Note that $A^TA = 12+6i$.
Now $B = b - [(A^T.b)/((A^T).A)]A = (3/5+i/5, 1) - [2/5-i6/5]/12+6i](-3-i, 2) = \cdots$
And here I am lost with my calculations...
Help, please?
I assume you want to use the Gram Schmidt orthonormalization method to construct two orthonormal vectors, given vectors $a$ and $b$. The inner product of complex vectors $v,w$ is not defined as $v^T\cdot w$, but as$^1$ $\overline{v^T}\cdot w$, i.e. you also need to take the complex conjugate of the first component. (Or depending on the book you use, conjugate the second component.) On wikipedia you can see how it is defined.
$a$ and $b$ are already orthogonal (calculate their inner product!), so $A=2a$ does not add simplicity. Now you only need to normalize $a$ and $b$.
The norm of $a$ is the square root of the inner product $\langle a,a\rangle$, and:
$\langle a,a\rangle = -\frac{3+i}{2}\cdot-\frac{3-i}{2} + 1\cdot 1 = \frac{10}{4}+1=\frac{7}{2}$, so your first vector is normalized if you divide it by $\sqrt{7/2}$.
I think you can do the other inner product yourself, now you know the correct definition of the standard inner product :)
$1$: Actually, the transposing of a vector is not really the standard general way to define the inner product, but as you were already using it, I decided to just roll with it instead of touching a running machine.