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

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 ?

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 :)?

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 $$D=0$$.

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.

• thank you for your advise , I think I have mistaken the gauss elimination operation as an orthogonal transform. – Mr.Guo May 21 '19 at 0:40
• as your advice, I plan to modify my method by : $Q*\begin{bmatrix}A & B\\ 0 & C \\ D & E\end{bmatrix} = \begin{bmatrix} H & F \\ 0 & G \\ 0 & 0 \end{bmatrix} \approx \begin{bmatrix} A & F \\ 0 & G \\ 0 & 0 \end{bmatrix}$ what's your opinion ? actually I'm planning to achieve the schmidt-kalman filter by qr method, in skf method , the covariance is updated but some block is not changed, maybe I could apply qr decom while keep A not changed, I think they are equivalent operations , right ? – Mr.Guo May 21 '19 at 0:51
• also for the second case , like : $Q* \begin{bmatrix} D & 0 & E \\ B & A & C \\ 0 & 0 & F \end{bmatrix} = \begin{bmatrix} G & H & I \\ 0 & L & J \\ 0 & 0 & K \end{bmatrix} \approx \begin{bmatrix} G & H & I \\ 0 & A & J \\ 0 & 0 & K \end{bmatrix}$ – Mr.Guo May 21 '19 at 0:53
• @Mr.Guo I can't help you much with the Schmidt-Kalman filter, but regarding the QR decomposition, as long as you keep in mind the properties of orthogonal matrices, it should be fine. That is, $H$ and $L$ can't be equal to $A$, and theirs columns must account for the norm of $D$ and $H$ respectively. – Jean-Claude Arbaut May 21 '19 at 4:52
• ok thank you any way :) – Mr.Guo May 22 '19 at 5:28