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(a) Construct an orthonormal basis of the space $R^3$ satisfying the requirment of the Gram-Schmidt prodcure from the basis

$v_{1}=(-3,4,0)$ , $v_{2}=(5,10,-24)$ , $v_{3}=(0,0,1)$

(b) Given that $w_{3}=v_{3}-\langle v_{3}, e_{2}\rangle e_{2}-\langle v_{3}, e_{1}\rangle e_{1}$ Deduce from this definition that $\langle w_{3}, {e_2}\rangle=\langle w_{3}, {e_1}\rangle=0$

Solution:

(a) $w_{3}=v_{3}-\langle v_{3}, e_{2}\rangle e_{2}-\langle v_{3}, e_{1}\rangle e_{1}$

$w_{3}=(0,0,1)+\dfrac{12}{13}(\dfrac{4}{13},\dfrac{3}{13},-\dfrac{12}{13})-0$

$w_{3}=(\dfrac{48}{169},\dfrac{36}{169},1-\dfrac{144}{169})$

$w_{3}=(\dfrac{48}{169},\dfrac{36}{169},\dfrac{25}{169})$

Hence $$||w_{3}||=\dfrac{5}{13}$$

Thus $$e_{3}=\dfrac{13}{5}w_{3}=(\dfrac{48}{169},\dfrac{36}{169},\dfrac{25}{169})$$

While i understand how to use the Gram -schmidt procedure to compute values, i dont quite underdtand the last part of the solution and how it links back to the question as well as part (b) which requires a deduction from the definitions. Cohld anyone explain. Thanks

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suppose that you've found the orthonormal basis $e_1,e_2,e_3$ from the Gram-schmidt process so for $1\le i,j\le 3$ we have $\langle e_i,e_j\rangle=0$ and $\langle e_i,e_i\rangle=1$ because the basis is an orthonormal one so we have:
$$\langle w_{3}, {e_2}\rangle=\langle v_{3}-\langle v_{3}, e_{2}\rangle e_{2}-\langle v_{3}, e_{1}\rangle e_{1},e_2\rangle=\langle v_3,e_2\rangle-\langle v_3,e_2\rangle\langle e_2,e_2\rangle-\langle v_3,e_1\rangle\langle e_1,e_2\rangle=\langle v_{3}, e_{2}\rangle-\langle v_{3}, e_{2}\rangle=0$$
and we have the same procedure for proving that $\langle w_{3}, e_{1}\rangle=0$
Now if you don't know how to calculate $e_1,e_2,e_3$ from the Gram-schmidt process, tell me to write it down for you but in order to answer the section (b) you don't need the values of $e_1,e_2,e_3$
you don't even need to find the value of $w_3$ as you did in your solution. You just have to use the definition of an orthonormal basis.
The definition is if $e_1,e_2,\dots,e_n$ is an orthonormal basis for an n-dimensional vector space found by the Gram-schmidt process then for $1\le e_1,e_2,\dots,e_n\le n$ we have $\langle e_i,e_j\rangle=0$ and $\langle e_i,e_i\rangle=1$ meaning that the vectors of an orthonormal basis are two-by-to perpendicular to each other having norm 1.

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