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Math Help - inner product , norm

  1. #1
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    inner product , norm

    Can someone please explain how the

    norm ||(3,4)|| =  \sqrt{69}

    I dont understand the working,

     \sqrt{<(4,3),(4,3)>}  = \sqrt{48-12-12+45} = \sqrt{69}



    ?

    and also how
    <(1,1),(2,1)> = 6+1-2-5 =0

    thank you.
    Attached Thumbnails Attached Thumbnails inner product , norm-screen-shot-2014-06-09-10.12.58.png  
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  2. #2
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    Re: inner product , norm

    Actually I worked it,

    you just the formula given for B

    3x1y1-x1y2-x2y1+5x2y2
    Last edited by Tweety; June 9th 2014 at 02:39 AM.
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  3. #3
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    Re: inner product , norm

    BY DEFINITION:

    $B((x_1,x_2),(y_1,y_2)) = 3x_1y_1 - x_1y_2 - x_2y_1 + 5x_2y_2$

    and $\|(x_1,x_2)\|_B \stackrel{\text{def}}{=} \sqrt{B((x_1,x_2,x_1,x_2))}$

    $= \sqrt{3(x_1)^2 - (x_1x_2) - (x_2x_1) + 5(x_2)^2} = \sqrt{3(x_1)^2 - 2(x_1x_2) + 5(x_2)^2}$.

    So, if $x_1 = 4$ and $x_2 = 3$, we have:

    $\|(4,3)\|_B = \sqrt{3(16) - 2(12) + 5(9)} = \sqrt{48 - 24 + 45} = \sqrt{69}$.

    Also by definition, $(x_1,x_2),(y_1,y_2)$ are orthogonal with respect to $B$, if:

    $B((x_1,x_2),(y_1,y_2)) = 0$. All we need to do is compute this:

    $B(1,1),(2,-1)) = 3(1\ast2) - (1\ast(-1)) - (1\ast2) + 5(1\ast(-1)) = 3(2) - (-1) - 2 + 5(-1) = 6 + 1 - 2 - 5 = 0$

    *****************

    If you prefer, we may define $B((x_1,x_2),(y_1,y_2))$ as:

    $\begin{bmatrix}x_1&x_2 \end{bmatrix} \begin{bmatrix} 3&-1\\-1&5 \end{bmatrix} \begin{bmatrix}y_1\\y_2 \end{bmatrix}$,

    the computations remain the same.
    Thanks from Tweety
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