Question

Consider a QRfactorization

M=QRShow that $\mathit{R}\mathbf{=}{\mathit{Q}}^{\mathbf{T}}\mathit{M}$.

Short answer

$R=Q^{T}M$

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 prove that$R=Q^{T}M$

Since, the column of the matrix Q is orthonormal and therefore it gives,

$$\begin{gathered} Q{Q}^T=l \\ =Q^{T}Q \end{gathered}$$

Thus, it is written as,

$$\begin{gathered} M=QR \\ \Rightarrow Q^{T}M=Q^{Q}R \\ \Rightarrow Q^{T}M=IR \\ \Rightarrow R=Q^{T}M \end{gathered}$$

Hence, $R=Q^{T}M$proved.