how to apply qr decomposition while keep top left block un-change?

Question

I have a matrix which is already up-triangular like :

$\begin{bmatrix}A & B\\ 0 & C\end{bmatrix}$

in which A and C is up-triangular block, now I add a block row then get :

$\begin{bmatrix}A & B\\ 0 & C \\ D & E\end{bmatrix}$

I want to apply a qr decomposition to keep this matrix up-triangular while keep A is not changed:

$ Q*\begin{bmatrix}A & B\\ 0 & C \\ D & E\end{bmatrix} = \begin{bmatrix} A & F \\ 0 & G \\ 0 & 0 \end{bmatrix}$

I know this could be achieved by gauss elimination method, because A is already up-triangular, we can modify the D col by col according the data from A, this could also keep A not changed.

but col by col operation is not computation efficient. I know the givens rotation or householder reflection is the common method for QR, but I think they can't keep A not changed, so any more efficient method than gauss elimination ?

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and I have an advance question, I have an up-triangular matrix like, which means a linear solve system :

$ \begin{bmatrix} A & B &C \\ 0 & D & E \\ 0 & 0 & F \end{bmatrix} $

A,D,F is up-triangular block. with some variables permute I can get another linear system:

$ \begin{bmatrix} D & 0 & E \\ B & A & C \\ 0 & 0 & F \end{bmatrix} $

I want apply a qr decomposition while keep A not changed, which mean:

$ Q* \begin{bmatrix} D & 0 & E \\ B & A & C \\ 0 & 0 & F \end{bmatrix} = \begin{bmatrix} G & H & I \\ 0 & A & J \\ 0 & 0 & K \end{bmatrix}$

I think even gauss elimination can't achieve this purpose. is there any solution :)?

Answer 1

First case

Note that multiplication by an orthogonal matrix preserves the $2$-norm of vectors. It's therefore impossible to achieve what you want unless $D=0$.

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Second case

Likewise, you must have $H=0$.

Now assume further that $A$ is regular (since it's upper triangular, this means it has no zero on the diagonal). Then, cut $Q$ in blocks so that $Q^T$ matches the block structure of the other two matrices.

Then multiplying $Q$ by the middle block-column, $Q_{12}A=0$, $Q_{22}A=A$, $G_{32}A=0$, hence $Q_{12}=0, Q_{22}=I, Q_{32}=0$.

Since the column vectors of $Q$ are orthogonal, you must also have $Q_{21}=0$ and $Q_{23}=0$. Then multiplying $Q$ by the first block column, you must have $B=0$, and multiplying by the last block-column, $C=J$.

But then, the matrix to triangularize is already upper triangular, so you may as well pick $Q=I$.

Note that when $A=0$ (it's then not regular anymore), it's possible to achieve the $QR$ decomposition.