Chapter 5.3, Problem 67E

Consider a subspace V of ${ℝ}^n$ with a basis ${\stackrel{\to }{v}}_1,\mathrm{...},{\stackrel{\to }{v}}_m$ ; supposewe wish to find a formula for the orthogonal projection onto V. Using the methods we havedeveloped thus far, we can proceed in two steps:We use the Gram-Schmidt process to construct anorthonormal basis ${\stackrel{\to }{v}}_1,\mathrm{...},{\stackrel{\to }{v}}_m$ of V, and then weuse Theorem 5.3.10: The matrix of the orthogonal projection is $Q{Q}^T$ , where $Q=\left[{\stackrel{\to }{u}}_1\cdots {\stackrel{\to }{u}}_m\right]$ .
In this exercise we will see how we can write the matrix of the projection directly in terms of the basis
${\stackrel{\to }{v}}_1,\mathrm{...},{\stackrel{\to }{v}}_m$ and the matrix $A=\left[{\stackrel{\to }{v}}_1\mathrm{...}{\stackrel{\to }{v}}_m\right]$ .
(This issue will be discussed more thoroughly in Section 5.4; see Theorem 5.4.7.)
Since ${\text{proj}}_V\stackrel{\to }{x}$ is in V. we can write ${\text{proj}}_V\stackrel{\to }{x}=c_1{\stackrel{\to }{v}}_1+\cdots +c_m{\stackrel{\to }{v}}_m$ for some scalars $c_1,\mathrm{...}{c}_m$ , yet to be determined. Now $\stackrel{\to }{x}-pro{j}_V\left(\stackrel{\to }{x}\right)=\stackrel{\to }{x}-c_1{\stackrel{\to }{v}}_1-\cdots c_m{\stackrel{\to }{v}}_m$ is orthogonal to V, meaning that ${\stackrel{\to }{v}}_i\cdot \left(\stackrel{\to }{x}-c_1{\stackrel{\to }{v}}_1-\cdots -c_m{\stackrel{\to }{v}}_m\right)=0$ for $i=1,\mathrm{...},m$ .
a. Use the equation ${\stackrel{\to }{v}}_i\cdot \left(\stackrel{\to }{x}-c_1{\stackrel{\to }{v}}_1-\cdots -c_m{\stackrel{\to }{v}}_m\right)=0$ to show that $A^{T}A\stackrel{\to }{c}=A^T\stackrel{\to }{x}$ , where $\stackrel{\to }{c}=\left[\begin{array}{l}{c}_1\\ ⋮\\ c_m\end{array}\right]$ .
b. Conclude that $\stackrel{\to }{c}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\to }{x}$ and ${\text{proj}}_{\text{V}}\stackrel{\to }{x}=A\stackrel{\to }{c}=A{\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\to }{x}$ .