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Suppose I have certain independent vectors, say $\lvert V_1\rangle$ and $\lvert V_2\rangle$, which span a 2-dimensional subspace of a given Complex Vector Space on which inner product is defined, how is the standard Gram Schmidt Process extended?

Even though StackExchange has answers to related questions, I have a problem with how exactly the method works. Following the process, we get $$\lvert v_1\rangle = \frac{\lvert V_1 \rangle}{\sqrt{\langle V_1 \rvert V_1 \rangle}}$$ and $$\lvert V_2' \rangle = \lvert V_2 \rangle - \langle V_2 \rvert v_1 \rangle \lvert v_1 \rangle$$ and $$\lvert v_2 \rangle = \frac{\lvert V_2 \rangle - \langle V_2 \rvert v_1 \rangle \lvert v_1 \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}$$Now $\lvert v_1 \rangle$ and $\lvert v_2 \rangle$ form an orthonormal basis for the given subspace. If this is true, then $\langle v_1 \rvert v_2 \rangle$ should be equal to 0.

But, $$\langle v_1 \rvert v_2 \rangle = \langle v_1 \rvert \left(\frac{\lvert V_2 \rangle - \langle V_2 \rvert v_1 \rangle \lvert v_1 \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}\right)$$ Since $\langle v_1 \rvert v_1 \rangle = 1$, $$\langle v_1 \rvert v_2 \rangle = \frac{\langle v_1 \rvert V_2 \rangle - \langle V_2 \rvert v_1 \rangle}{\sqrt{\langle V_2 \rvert V_2 \rangle}}$$ $$\langle v_1 \rvert v_2 \rangle = \frac{\langle v_1 \rvert V_2 \rangle - {\langle v_1 \rvert V_2 \rangle}^*}{\sqrt{\langle V_2 \rvert V_2 \rangle}}$$ But the final equation is not necessarily zero for a complex vector space. Am I going wrong somewhere?

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  • $\begingroup$ Yes, but why don't you define $\lvert v_2 \rangle = \frac{\lvert V_2 \rangle - \langle v_1 \rvert V_2 \rangle \lvert v_1 \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}$? $\endgroup$ – Raskolnikov Jan 28 at 22:16
  • $\begingroup$ @Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method. $\endgroup$ – Ajay Shanmuga Sakthivasan Jan 28 at 22:30
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The problem lies in your definition of $\lvert v_2\rangle$. It should be$$\lvert v_2 \rangle = \frac{\lvert V_2'\rangle - \langle V_2'\rvert v_1 \rangle \lvert v_1 \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}.$$Using this definition, you will get that, indeed, $\langle v_1|v_2\rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$\lvert v_2 \rangle = \frac{\lvert V_2'\rangle - \langle V_2'\rvert v_1 \rangle^*\lvert v_1 \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}.$$

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  • $\begingroup$ No, but $\lvert v_2 \rangle$ would be $\frac{\lvert V_2' \rangle}{\sqrt{\langle V_2' \rvert V_2' \rangle}}$, where I have written $\lvert V_2' \rangle$ using the second equation. Even if I take $\lvert v_2 \rangle$ as given in your answer, I would end up in the same trouble. $\endgroup$ – Ajay Shanmuga Sakthivasan Jan 28 at 22:27
  • $\begingroup$ I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$\lvert V_2' \rangle=\lvert V_2\rangle-\langle V_2 \rvert v_1 \rangle^*\lvert v_1 \rangle$$and all will be fine. $\endgroup$ – José Carlos Santos Jan 28 at 22:48
  • $\begingroup$ This would indeed work, Thank you. $\endgroup$ – Ajay Shanmuga Sakthivasan Jan 28 at 23:20

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