If we assume that the definition of the inner product of the normalized vectors ${ˆ u}_{1}$ $hatu_1$ and ${ˆ u}_{2}$ $hatu_2$ is $2 u_{1}^{2} + u_{2}^{2}$ $2u_1^2 + u_2^2$ (these are vector components), how do you use the Gram-Schmidt process to generate orthonormal vectors from ${\to v}_{1}$ $vecv_1$ and ${\to v}_{2}$ $vecv_2$ ?

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If we assume that the definition of the inner product of the normalized vectors ${ˆ u}_{1}$ $hatu_1$ and ${ˆ u}_{2}$ $hatu_2$ is $2 u_{1}^{2} + u_{2}^{2}$ $2u_1^2 + u_2^2$ (these are vector components), how do you use the Gram-Schmidt process to generate orthonormal vectors from ${\to v}_{1}$ $vecv_1$ and ${\to v}_{2}$ $vecv_2$ ?

${\to v}_{1} = (2, 1)$ $vecv_1 = (2,1)$
${\to v}_{2} = (5, - 7)$ $vecv_2 = (5,-7)$

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Answer 1 · 2017-07-31T19:12:39

See below.

Explanation:

Following the Gram-Schmidt orthonormalization process, given ${{\to u}_{1}, {\to u}_{2}}$ ${vec u_1, vec u_2}$ we begin by one of then, for instance ${\to u}_{1}$ $vec u_1$ and define

${\to v}_{1} = \frac{{\to u}_{1}}{{∥ ∥ {\to u}_{1} ∥ ∥}_{q}}$ $vec v_1 = (vec u_1)/norm(vec u_1)_q$

and then

${\to v}_{2} = \frac{{\to u}_{2} - {⟨ {\to u}_{2}, {\to v}_{1} ⟩}_{q} {\to v}_{1}}{{∥ ∥ ∥ {\to u}_{2} - {⟨ {\to u}_{2}, {\to v}_{1} ⟩}_{q} {\to v}_{1} ∥ ∥ ∥}_{q}}$ $vec v_2 = (vec u_2 - << vec u_2,vec v_1 >>_q vec v_1)/norm( vec u_2 - << vec u_2,vec v_1 >>_q vec v_1)_q$

so if

${(vec u_1=(2,1)),(vec u_2=(5,-7)):}$

we obtain

${(vec v_1 =(2/3, 1/3) ),(vec v_2 =(sqrt 2/3, -(2 sqrt[2])/3) ):}$

NOTE: Here $<< cdot, cdot >>_q$ applied to two vectors $vec u, vec v$ gives $<< vec u, q cdot vec v >>$ and

$norm(vec u)_q = sqrt(<< vec u, q cdot vec u >>)$ with $q$ a definite positive matrix.

In the present case $q = ((2,0),(0,1))$

Answer 2 · 2017-07-31T20:17:48

I did your question from the beginning and got:

$hatu_1 = (2/3,1/3)$

$hatu_2 = (1/(3sqrt2), -4/(3sqrt2))$

The Gram-Schmidt process for two vectors first involves orthogonalizing them:

$vecu_1 = vecv_1$ , $" "" "" "" "" "" "" "hatu_1 = vecu_1/||vecu_1||$

$vecu_2 = vecv_2 - "proj"_(vecu_1)vecv_2$ , $" "" "hatu_2 = vecu_2/||vecu_2||$

And if you had more than two vectors $vecu_i$ , you would match the subscript of $vecu_i$ with $vecv_i$ , then sequentially subtract the projection of the current $vecv_i$ (highest index $i$ ) onto the previous $vecu_i$ 's.

You have been given

$vecv_1 = (2, 1)$
$vecv_2 = (5, -7)$

Now, you already have $hatu_1$ . Due to the definition of your inner product, the norm is not the usual definition (difference highlighted in red):

$|| vecu ||^2 = << vecu, vecu >> = color(red)(2)u_1u_1 + u_2u_2$

$=> color(blue)(hatu_1) = ((2", "1))/(sqrt(color(red)(2) xx (2)^2 + 1^2)) = ulcolor(blue)((2/3, 1/3)" ")$ ,

which [indeed has a norm of $1$ under YOUR inner product definition.](http://www.wolframalpha.com/input/?i=sqrt%282+*+%282%2F3%29^2+%2B+%281%2F3%29^2%29)

To proceed, we define the projection of $bb(vecv_2)$ onto $bb(vecu_1)$ :

$"proj"_(vecu_1)vecv_2 = (<< vecv_2, vecu_1 >>)/(<< vecu_1, vecu_1 >>) vecu_1$

How I remember it is that the vector that is projected (mapped), $vecv_2$ , is the only different term in the projection definition.

We continue by finding the inner product of $vecv_2$ with $vecu_1$ , using YOUR definition, which is again, not quite the dot product of the vectors:

$<< vecv_2, vecu_1 >> = color(red)(2)v_(21)u_(11) + v_(22)u_(12)$

$= color(red)(2) xx 5 cdot 2 + -7 cdot 1$

$= 13$

And the inner product of $vecu_1$ with itself is the norm squared:

$<< vecu_1, vecu_1 >> = ||vecu_1||^2 = 3^2 = 9$

As a result, the projection of $vecv_2$ onto $vecu_1$ is:

$"proj"_(vecu_1)vecv_2 = 13/9 cdot (2, 1)$

$= (26/9, 13/9)$

and so, the vector $vecu_2$ would be found as:

$color(red)(vecu_2) = vecv_2 - "proj"_(vecu_1)vecv_2$

$= (5, -7) - (26/9, 13/9)$

$= (45/9, -63/9) - (26/9, 13/9)$

$= color(red)((19/9, -76/9)" ")$

Note that this is NOT normalized yet. The normalization of this is then:

$color(blue)(hatu_2) = (vecu_2)/(||vecu_2||)$

$= ((19/9", "-76/9))/sqrt(color(red)(2) xx (19/9)^2 + (-76/9)^2)$

$= (19/(19 sqrt2/3 cdot 9)", "-76/(19 sqrt2/3 cdot 9))$

$= ulcolor(blue)((1/(3sqrt2)", "-4/(3sqrt2))" ")$

And indeed this has a magnitude of $1$ under this inner product definition.

Lastly, to check whether we are correct, we should see if the inner product of $hatu_1$ and $hatu_2$ is $0$ (for orthogonality only). We already know these vectors are properly normalized in this particular inner product space.

$0 stackrel(?" ")(=) << hatu_1, hatu_2 >>$

$= color(red)(2) xx 2/3 cdot 1/(3sqrt2) + 1/3 cdot -4/(3sqrt2)$

$= 4/(9sqrt2) + (-4/(9sqrt2)) = 0$ $color(blue)(sqrt"")$

So, this should be correct! (If you wish, you can check that $<< hatu_1, hatu_1 >> = << hatu_2, hatu_2 >> = 1$ , which is again, $color(red)(2)u_(11)^2 + u_(12)^2$ , and $color(red)(2)u_(21)^2 + u_(22)^2$ , respectively.)

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${\to v}_{1} = (2, 1)$ $vecv_1 = (2,1)$
${\to v}_{2} = (5, - 7)$ $vecv_2 = (5,-7)$

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vecv_1 = (2,1)→v1=(2,1)vecv_1 = (2,1) vecv_2 = (5,-7)→v2=(5,−7)vecv_2 = (5,-7)

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${\to v}_{1} = (2, 1)$ $vecv_1 = (2,1)$
${\to v}_{2} = (5, - 7)$ $vecv_2 = (5,-7)$