Orthogonalizing vectors. Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is

Question

Orthogonalizing vectors. Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if b 6= 0. (Give a formula for the scalar γ.) Roughly speaking, we can always subtract a multiple of a vector from another one, so that the result is orthogonal to the original vector. This is called orthogonalization, and is a basic idea used in the Gram-Schmidt algorithm.

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Kylie 2 weeks 2021-11-17T15:51:39+00:00 1 Answer 0 views 0

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    2021-11-17T15:53:15+00:00

    Answer:

    \\ \gamma= \frac{a\cdot b}{b\cdot b}

    Step-by-step explanation:

    The question to be solved is the following :

    Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if b \neq 0. Recall that given two vectors a,b  a⊥ b if and only if a\cdot b =0 where \cdot is the dot product defined in \mathbb{R}^n. Suposse that b\neq 0 . We want to find γ such that (a-\gamma b)\cdot b=0. Given that the dot product can be distributed and that it is linear, the following equation is obtained

    (a-\gamma b)\cdot b = 0 = a\cdot b - (\gamma b)\cdot b= a\cdot b - \gamma b\cdot b

    Recall that a\cdot b, b\cdot b are both real numbers, so by solving the value of γ, we get that

    \gamma= \frac{a\cdot b}{b\cdot b}

    By construction, this γ is unique if b\neq 0, since if there was a \gamma_2 such that (a-\gamma_2b)\cdot b = 0, then

    \gamma_2 = \frac{a\cdot b}{b\cdot b}= \gamma

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