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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

"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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.