inner product , norm

• Jun 9th 2014, 02:19 AM
Tweety
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.
• Jun 9th 2014, 03:14 AM
Tweety
Re: inner product , norm
Actually I worked it,

you just the formula given for B

3x1y1-x1y2-x2y1+5x2y2
• Jun 9th 2014, 03:24 AM
Deveno
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.