# Math Help - need to know insight concept or underlaying concept of Cauchy–Schwarz inequality

1. ## need to know insight concept or underlaying concept of Cauchy–Schwarz inequality

CAN ANYONE PLEASE TELL ME ABOUT
Cauchy–Schwarz inequality
1. WHAT IT MEANS ACTUALLY I KNOW FORMAL DEFINITION BUT I WANA INSIGHT CONCEPT WHY PRODUCT OF INDIVIDUAL VECTORS IS ALWAYS> AND EQUAL THAN INNER PRODUCT

2. SPECIFY THE CASE WHEN PRODUCT OF MAGNITUDES OF IND.VECTORS IS = MAG OF INNER PRODUCT
MAY THIS HELPS ME TO GET INSIGHT CONCEPT

I AM NEW TO THESES FORUMS IF ANYONE CAN EMAIL ME IT WILL BE GREAT FOR ME...
THANKX.
E-MAIL ADRESS:
junaidanwar194@gmail.com

2. ## Re: need to know insight concept or underlaying concept of Cauchy–Schwarz inequality

Suppose you have two UNIT vectors $\hat{v}, \hat{w}$ in some inner product space (V, <,>).

Then $\vec{z} = <\hat{v}, \hat{w}> \hat{w}$ is the component of $\hat{v}$ in the $\hat{w}$ direction.

(Recall that $\hat{v} = \vec{z} + (\hat{v} - \vec{z})$, decomposes $\hat{v}$ into parallel and perpendicular components w/resp to $\hat{w}$:

$<(\hat{v} - \vec{z}), \hat{w}> = <\hat{v}, \hat{w}> - <\vec{z}, \hat{w}> = <\hat{v}, \hat{w}> - <<\hat{v}, \hat{w}> \hat{w}, \hat{w}>$

$= <\hat{v}, \hat{w}> - <\hat{v}, \hat{w}><\hat{w}, \hat{w}> = <\hat{v}, \hat{w}>( 1 - <\hat{w}, \hat{w}>) = 0$.)

Conceptually, should $\hat{v}$, some generic unit vector, have a greater component in the $\hat{w}$ direction than $\hat{w}$ itself does?

No. Conceptually, each unit vector "should have the greatest magnitude among all unit vector in its own direction." Right?

OK. So that means we expect.... $\lVert \vec{z} \rVert \le \lVert \hat{w} \rVert = 1$, with equality iff $\hat{v} = \pm \hat{w}$.

Thus we expect.... $\lVert <\hat{v}, \hat{w}> \hat{w} \rVert \le 1$, with equality iff $\hat{v} = \pm \hat{w}$.

Thus we expect.... $\lvert <\hat{v}, \hat{w}> \rvert \lVert \hat{w} \rVert \le 1$, with equality iff $\hat{v} = \pm \hat{w}$.

Thus we expect.... $\lvert <\hat{v}, \hat{w}> \rvert \le 1$, with equality iff $\hat{v} = \pm \hat{w}$.

Thus we expect inner products of *unit* vectors to take values in [-1,1], with it being $\pm1$ only when one vector is $\pm1$ the other.

Now, suppose $\vec{a}$ and $\vec{b}$ are any two non-zero vectors (CS obviously holds if one of the vectors is $\vec{0}$). What do we expect?

Consider the proceeding in light of their unit vectors $\frac{\vec{a}}{\lVert \vec{a} \rVert}$ and $\frac{\vec{b}}{\lVert \vec{b} \rVert}$:

We expect.... $\left\lvert \left<\frac{\vec{a}}{\lVert \vec{a} \rVert}, \frac{\vec{b}}{\lVert \vec{b} \rVert}\right> \right\rvert \le 1$, with equality iff $\frac{\vec{a}}{\lVert \vec{a} \rVert} = \pm \frac{\vec{b}}{\lVert \vec{b} \rVert}$, i.e. iff $\vec{a} = \left( \pm \frac{\lVert \vec{a} \rVert }{\lVert \vec{b} \rVert } \right)\vec{b}$.

So we expect.... $\lvert <\vec{a}, \vec{b}> \rvert \le \lVert \vec{a} \rVert \ \lVert \vec{b} \rVert$, with equality iff $\vec{a}$ = some multiple of $\vec{b}$.

Thus we expect the Cauchy-Schwarz inequality to hold - because we expect inner products of unit vectors to take values in [-1,1].

That's its conceptual meaning to me.