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Hope you're doing well.

Well I'm stuggling to understand the proof of Cauchy-Schwarz inequality, especially with the following equation: $$ \frac{1}{\|\mathbf{v}\|^{2}}\|\| \mathbf{v}\left\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\right\|^{2}=\|\mathbf{u}\|^{2}\|\mathbf{v}\|^{2}-|\langle\mathbf{u}, \mathbf{v}\rangle|^{2} $$

As it says in wikipedia: if you simplify the right side of the equation you'll get the left side easily. I tried and it doesn't work for me.

Thanks in advance for your answers.

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    $\begingroup$ note that $\Vert x\Vert^2 = \langle x,x\rangle$ $\endgroup$
    – lmaosome
    Jan 24, 2021 at 22:13
  • $\begingroup$ Is the statement you've written true for all inner products or just for the dot product. Thanks for your response though. $\endgroup$
    – Saad Jlil
    Jan 25, 2021 at 8:30
  • $\begingroup$ any inner product defines a norm. but be aware that there are norms which are not induced by inner products $\endgroup$
    – lmaosome
    Jan 25, 2021 at 13:19

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Note that\begin{align}\bigl\|\| \mathbf{v}\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\bigr\|^2&=\|\mathbf v\|^4\|\mathbf u\|^2-2\|\mathbf v\|^2\langle\mathbf u,\mathbf v\rangle^2+\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\|\mathbf v\|^2\\&=\|\mathbf v\|^2\bigl(\|\mathbf v\|^2\|\mathbf u\|^2-\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\bigr).\end{align}Now, divide both sides by $\|\mathbf v\|^2$ and you will get your equality.

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  • $\begingroup$ Is the following statement true (because the inner product doesn't have to be a dot product): $\langle\mathbf{u}, \mathbf{u}\rangle=\|\mathbf{u}\|^{2}$ $\endgroup$
    – Saad Jlil
    Jan 25, 2021 at 8:27
  • $\begingroup$ Yes, it is true. That's how the norm is defined in an inner product space. $\endgroup$
    – José Carlos Santos
    Jan 25, 2021 at 8:30
  • $\begingroup$ Ok thanks a lot $\endgroup$
    – Saad Jlil
    Jan 26, 2021 at 17:29

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