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.

Answer 1

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

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