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I need a proof for the Cauchy-Schwarz inequality on complex numbers, i.e.

$$|{(a_1b_1 + a_2b_2+...+a_nb_n)}|^2\leq(|a^2_1|+|a^2_2|+...+|a^2_n|) (|b^2_1|+|b^2_2|+...+|b^2_n|),$$ where each $a_i,b_i\in\mathbb{C}$. I thought of proving the $real$ and the $imaginary$ parts separately and then summing them up, but that will not work because on doing so the $real$ nos. formed by multiplying the $imaginary$ part of both numbers will not be considered. Please help me prove it.

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  • $\begingroup$ There is indeed a cauchy-schwarz inequality for complex number. But it is not of your current form. Both sides of your inequalities might be complex number currently and makes comparison meaningless. To fix this, you might like to use some norm (or absolute value). $\endgroup$ May 9, 2016 at 6:26
  • $\begingroup$ @SiongthyeGoh sorry, i have made the required changes.... $\endgroup$ May 9, 2016 at 13:22
  • $\begingroup$ @GEdgar if i put in your substitution, i get zero on both sides. Actually, equality holds if $\frac{a_i}{b_i}=\frac{a_j}{b_j}$. thus your case results in equality $\endgroup$ May 9, 2016 at 13:36
  • $\begingroup$ Now try to use Kavi's information to come up with the proof. $\endgroup$
    – GEdgar
    May 9, 2016 at 13:44
  • $\begingroup$ @GEdgar, yes Kavi"s statement is correct and that is exactly what i am unable to prove $\endgroup$ May 9, 2016 at 13:46

2 Answers 2

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There is no accepted definition of inequalities for complex numbers. The correct statement of Cauchy - Schwartz inequality is the the following: $\left\vert\ a_1\bar{b_1}+…+a_n\bar{b_n}\right\vert\leq \left(|a_1|^2+...+|a_1|^2\right) \left(|b_1|^2+...+ |b_1|^2 \right)$ and this follows immediately from the real case.

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  • $\begingroup$ I had forgotten to add the modulus... $\endgroup$ May 9, 2016 at 13:23
  • $\begingroup$ Notice that you must use $|a_1|^2$ and so on. Without the modulus inside there, you can have $|a_1^2+\dots +a_n^2|=0$ equal to zero even if the $a_j$ are not zero. $\endgroup$
    – GEdgar
    May 9, 2016 at 13:26
  • $\begingroup$ @GEdgar....yes, that is my mistake, ill correct it.... $\endgroup$ May 9, 2016 at 13:39
  • $\begingroup$ How does the |a_1^2 +......a_n^2|= 0 leads to anything wrong why we cannot have zero value? @GEdgar $\endgroup$ Apr 24, 2022 at 12:43
  • $\begingroup$ @Paracetamol... The problem with the original formulation of the question $${(a_1b_1 + a_2b_2+...+a_nb_n)}^2\leq(a^2_1+a^2_2+...+a^2_n)(b^2_1+b^2_2+...+b^2_n)$$ is that $(a^2_1+a^2_2+...+a^2_n) = 0$ is possible even if $(a_1b_1 + a_2b_2+...+a_nb_n) \ne 0$. $\endgroup$
    – GEdgar
    Apr 24, 2022 at 14:03
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The Cauchy-Schwarz inequality tells us that $ |\langle u, v \rangle| \leq |u| |v| $ in any inner product space where the norm is the norm induced by the inner product ($ |v| = \sqrt{\langle v, v \rangle} $). The standard inner product for complex-valued vectors is given by $ \langle v, w \rangle = v^H w $ where $ v^H $ is the conjugate transpose of $ v $ (take complex conjugates of each entry and then transpose), one can indeed verify that this is an inner product on $ \mathbb{C} $.

Now, the only thing left is to plug and chug. Let $ u = (u_1, u_2, \ldots, u_n) $ and $ v = (v_1, v_2, \ldots, v_n) $, then we have

$$ |\langle u, v \rangle| = |u^H v| = \left| \sum_{k=1}^{n} \bar{u_k} v_k \right| $$ $$ |u| |v| = \sqrt{ \sum_{k=1}^{n} \bar{u_k} u_k } \sqrt{ \sum_{k=1}^{n} \bar{v_k} v_k } = \sqrt{ \left(\sum_{k=1}^{n} |u_k|^2 \right) \left(\sum_{k=1}^{n} |v_k|^2 \right)} $$

Putting these expressions into the Cauchy-Schwarz inequality and squaring both sides then yields the desired result.

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