# Relation between the QR decomposition of matrix A and the QR of the augmented version of A

suppose I have a matrix $A_{1}\in C^{m*n}, m\geqslant n$ so the QR decomposition of $A_{1}$ is $A_{1}=Q_{1}*R_{1}$. Now, define the augmented matrix $A_{2}=\begin{bmatrix} A_{1}\\ I_{n*n} \end{bmatrix}$ where $I_{n*n}$ is $n*n$ identity matrix, so the QR decomposition of $A_{2}$ is $A_{2}=Q_{2}*R_{2}$.
Is there any relation between $Q_{1}$ and $Q_{2}$, $R_{1}$ and $R_{2}$, and especially between the diagonal elements of $R_{1}$ and $R_{2}$ knowing that $det(I_{n*n}+A_{1}^{H}*A_{1})=det(A_{2}^{H}*A_{2})$.

## 2 Answers

First consider the simple case, if $Q_1 = I$, then you want to find $Q_2$ and $R_2$ such that $Q_2 R_2 = \begin{bmatrix} R_1 \\ I \end{bmatrix}$. Split the columns of $Q_2^T$ into left and right halves such taht $Q_2^T = [U V]$. Note that $U^TU = V^TV = I$ and $U^TV = 0$. Then basically you want to find $R_2 = UR_1 + V$ which implies that $U^TR_2 = R_1$.

The more general case can be analyzed by considering the relation that $Q_2R_2 = \begin{bmatrix} Q_1R_1 \\ Q_1Q_1^T \end{bmatrix}$ and therefore $R_2 = UQ_1R_1 + V$ which leads to the result that $Q_1^TU^TR_2 = R_1$.

I think the relation is not that simple and here is why:

Since $A_1 = Q_1R_1$ we have $$A_2 = \begin{bmatrix}A_1\\ I\end{bmatrix} = \begin{bmatrix}Q_1 & 0\\0 & I\end{bmatrix}\begin{bmatrix}R_1\\I\end{bmatrix}.$$ Now the QR decomposition of $\begin{bmatrix}R_1\\I\end{bmatrix}$ can be done by $n(n-1)/2$ Givens rotations and I don't see any reason, why this should lead to a simple relation of $R_1$ and $R_2$.