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.

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 [tex]b \neq 0[/tex]. Recall that given two vectors a,b a⊥ b if and only if [tex]a\cdot b =0[/tex] where [tex]\cdot[/tex] is the dot product defined in [tex]\mathbb{R}^n[/tex]. Suposse that [tex]b\neq 0 [/tex]. We want to find γ such that [tex](a-\gamma b)\cdot b=0[/tex]. Given that the dot product can be distributed and that it is linear, the following equation is obtained

[tex](a-\gamma b)\cdot b = 0 = a\cdot b - (\gamma b)\cdot b= a\cdot b - \gamma b\cdot b[/tex]

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

[tex]\gamma= \frac{a\cdot b}{b\cdot b}[/tex]

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

[tex]\gamma_2 = \frac{a\cdot b}{b\cdot b}= \gamma[/tex]