Gram Schmidt Process with inner product $\langle z,w\rangle = 3(z_1)(\bar{w_1}) + 2(z_2)(\bar{w_2})+i(z_1)(\bar{w_2})-i(z_2)(\bar{w_1})$

We are given an example where we have $v_1 = (1,0)$ and $v_2=(0,1)$ on the complex plane with the inner product $\langle z,w\rangle = 3(z_1)(\bar{w_1}) + 2(z_2)(\bar{w_2})+i(z_1)(\bar{w_2})-i(z_2)(\bar{w_1})$. We are then given $\Vert v_1\Vert^2= 3$.

Why?

Also, how would one calculate the inner product $\langle v_2,u_1\rangle u_1$ to equal $\frac{1}{3}(-i)(1,0)$?

• $\|v_1\|^2=\left<v_1,v_1\right>$, so you just need to substitute $1$ and $0$ into the formula. For the second one you need to know what $u_1$ is. Is it $v_1$? – Arnaud Mortier Feb 16 '18 at 14:06
• $u_1$=$\frac{1}{2}(1,0)$ How do you substitute in 1 and 0 into the formula? There are no v's in the formula. – user532092 Feb 16 '18 at 14:20

The point is that when you write $v_1=(1,0)$, you use the same font as the coordinates when you write $\mathbf{w}=(w_1,w_2)$. That's what you find confusing I believe. You should be writing $$\mathbf{v}_1=(1,0)$$ and realise that this means $v_1=1$ and $v_2=0$.
From there you compute $\|\mathbf{v}_1\|^2=\left<\mathbf{v}_1,\mathbf{v}_1\right>$ by substituting the coordinates in your formula. Only the part $3z_1\bar{w}_1$ will be non-zero, and it will output $3$ as expected.