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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 $\frac{1}{3}(-i)(1,0)$?

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  • $\begingroup$ $\|v_1\|^2=\left<v_1,v_1\right>$, 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 $v_1$? $\endgroup$ – Arnaud Mortier Feb 16 '18 at 14:06
  • $\begingroup$ $u_1$=$\frac{1}{2}(1,0)$ How do you substitute in 1 and 0 into the formula? There are no v's in the formula. $\endgroup$ – user532092 Feb 16 '18 at 14:20
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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.

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  • $\begingroup$ Ok, that is what I was doing, and finally got the correct answer. I wrote the wrong value in for one of the components and thought I did not understand something! Thank you for your help. $\endgroup$ – user532092 Feb 16 '18 at 17:16
  • $\begingroup$ @user532092 You're welcome :) well done then. $\endgroup$ – Arnaud Mortier Feb 16 '18 at 17:17

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