Question

If A=QRis a QRfactorization, what is the relationship between ${\mathit{A}}^{\mathbf{T}}\mathit{A}$ and ${\mathit{R}}^{\mathbf{T}}\mathit{R}$?

Short answer

$A^{T}A=R^{T}R$

Step-by-step solution

QR factorization

Consider an n×mmatrix Mwith linearly independent column ${\stackrel{\mathbf{\perp }}{\mathbf{v}}}_1\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{\perp }}{\mathbf{v}}}_{\mathbf{m}}$.Then there exists an n×mmatrix Qwhose columns ${\stackrel{\mathbf{1}}{\mathbf{u}}}_1\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{1}}{\mathbf{u}}}_{\mathbf{m}}$are orthonormal and an upper triangular matrix Rwith positive diagonal entries such that.

M=QR

To find the relation between $A^{T}A\mathrm{and}{R}^{T}R$consider.

Since, the column of the matrix Q is orthonormal and therefore it has $Q{Q}^T=l=Q^{T}Q$. Thus, it is find as

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

Hence, the relation between $A^{T}A$and $R^{T}R$is $A^{T}A=R^{T}R$.