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I'm trying to prove Cauchy-Schwarz inequality in an complex inner product space when $\mathbf{x}=\lambda\mathbf y, \lambda\not=0$. but I don't know why the last step means equal:

$$\large|\langle\mathbf \lambda\mathbf y,\mathbf y\rangle|\le \lVert\lambda\mathbf y\rVert\lVert\mathbf y\rVert\\ \large\implies |\lambda||\langle\mathbf y,\mathbf y\rangle|\le|\lambda|\lVert\mathbf y\rVert^2\\ \large\implies|\langle\mathbf y,\mathbf y\rangle|\le\lVert\mathbf y\rVert^2.\ \ \ \ \ \ \ \ $$

Since $\lVert\mathbf y\rVert^2=\langle\mathbf y,\mathbf y\rangle,$ but why it's the same as $|\langle\mathbf y,\mathbf y\rangle|$?

My confusion came from the absolute value of complex number, $|a+bi|=\sqrt{a^2+b^2},$ and I was considering $\langle\mathbf y,\mathbf y\rangle$ be a complex number so I don't know why the equality $\langle\mathbf y,\mathbf y\rangle=|\langle\mathbf y,\mathbf y\rangle|$ will hold.

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  • $\begingroup$ That is how the norm is defined in an inner product space. $\endgroup$ – copper.hat Aug 15 '18 at 15:08
  • $\begingroup$ @copper.hat: But isn't that they're different: $||\mathbb y||:=\sqrt{\langle\mathbb y,\mathbb y\rangle},$ which there are double-vertical bars in the definition but $|\langle\mathbb y,\mathbb y\rangle|,$ where are single-vertical bars? $\endgroup$ – linear_combinatori_probabi Aug 15 '18 at 15:12
  • $\begingroup$ The inner product is required to satisfy $\langle y, y \rangle \ge 0$, so $\|y\|^2 = \langle y, y \rangle$ is well defined. Use $\mathbb{R}^n$ for intuition, here $\langle x, y \rangle = x^* y$. $\endgroup$ – copper.hat Aug 15 '18 at 15:14
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    $\begingroup$ I will write it again: Part of the definition of the inner product is that it must satisfy $\langle y, y \rangle \ge 0$, that is, if the same parameter is passed in both places the the result must be real & non negative. This is a definition. FOr complex finite dimensional vectors it is often defined as $y^*y$. Some authors use $y^T \bar{y}$. $\endgroup$ – copper.hat Aug 15 '18 at 15:52
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"Inner product" generally includes "positive definite" as an axiom, so $\langle \mathbf{y},\mathbf{y}\rangle$ is certainly nonnegative, so it equals its absolute value.

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  • $\begingroup$ Is this correct: since $\langle\mathbb y, \mathbb y\rangle$ will be a real number not complex number so the absolute value sign can be taken away and the right hand side of similar reasoning? $\endgroup$ – linear_combinatori_probabi Aug 15 '18 at 15:27
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    $\begingroup$ It will not only be real but positive (as long as $\mathbf{y}$ is nonzero): this is what it means that $\langle \cdot,\cdot\rangle$ is "positive definite." The absolute value sign can be removed because the absolute value of a positive number is the number itself. $\endgroup$ – Ben Blum-Smith Aug 15 '18 at 15:33

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