Question

The formula$\mathit{A}{\left(A^{T}A\right)}^{\mathbf{-}\mathbf{1}}$ for the matrix of an orthogonalprojection is derived in Exercise 67. Now considerthe QRfactorization of A, and express the matrix$\mathit{A}{\mathit{\left(}{\mathit{A}}^{\mathit{T}}\mathit{A}\mathit{\right)}}^{\mathit{-}\mathit{1}}{\mathit{A}}^{\mathit{T}}$in terms of Q.

Short answer

The equation$Q^{T}Q=I_m$ holds.

Step-by-step solution

Consider the theorem below.

The matrix of an orthogonal projection:Consider a subspace Vof${\mathit{R}}^{\mathbf{n}}$with orthonormal basis role="math" localid="1659500909794" ${\overrightarrow{\mathbf{u}}}_{\mathbf{1}},{\overrightarrow{\mathbf{u}}}_{\mathbf{2}},.......{\overrightarrow{\mathbf{u}}}_{\mathit{m}}$. The matrix Pof the orthogonal projection onto Vis

$\mathit{P}\mathbf{=}\mathit{Q}{\mathit{Q}}^{\mathbf{T}}$Where $\mathit{Q}\mathbf{=}\left[\begin{array}{ccc}|& |& |\\ {\overrightarrow{u}}_1& {\overrightarrow{u}}_2...& {\overrightarrow{u}}_m\\ |& |& |\end{array}\right]$.

Pay attention to the order of the factors ( $\mathit{Q}{\mathit{Q}}^{\mathbf{T}}$as opposed to ${\mathit{Q}}^{\mathbf{T}}\mathit{Q}$). Note that matrix Pis symmetric, since

$$\begin{gathered} {\mathit{P}}^{\mathbf{T}}\mathbf{=}{\left(Q{Q}^T\right)}^{\mathbf{T}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\left(Q^T\right)}^{\mathbf{T}}{\mathit{Q}}^{\mathbf{T}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{Q}{\mathit{Q}}^{\mathbf{T}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{P} \end{gathered}$$

So, if A = QR , then,

$$\begin{gathered} A{\left(A^{T}A\right)}^{-1}A=QR\left(R^T{Q}^{T}Q{R}^{-1}\right)R^T{Q}^T \\ =QR\left(R^T{R}^{-1}\right)R^T{Q}^T \\ =QR{R}^{-1}{\left(R^T\right)}^{-1}{R}^T{Q}^T \\ =Q{Q}^T \end{gathered}$$

As mentioned in the above theorem.

Hence, the equation $Q^{T}Q=I_m$holds since the columns of Q are orthonormal.