Results 1 to 2 of 2

Thread: need to know insight concept or underlaying concept of Cauchy–Schwarz inequality

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    RAWALPINDI
    Posts
    8

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

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

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

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

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

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

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

    Conceptually, should $\displaystyle \hat{v}$, some generic unit vector, have a greater component in the $\displaystyle \hat{w}$ direction than $\displaystyle \hat{w}$ 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.... $\displaystyle \lVert \vec{z} \rVert \le \lVert \hat{w} \rVert = 1$, with equality iff $\displaystyle \hat{v} = \pm \hat{w}$.

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

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

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

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


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

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

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

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

    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.
    Last edited by johnsomeone; Oct 1st 2012 at 09:25 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Oct 1st 2012, 06:58 AM
  2. Cauchy-Schwarz inequality
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Feb 9th 2011, 04:25 AM
  3. Cauchy-Schwarz Inequality
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Jun 27th 2010, 12:28 PM
  4. cauchy schwarz inequality
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Jan 30th 2010, 11:09 PM
  5. Cauchy-Schwarz Inequality
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 4th 2008, 05:20 AM

Search Tags


/mathhelpforum @mathhelpforum