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So I'm trying to factorize this matrix.

$A= \left( \begin{array}{ccc} 3 & 0 \\ 0 & 3 \\ 4 & 0\\ 0&4 \end{array} \right)$

So I need to remove the 4 at $a_{1,3}$, however I'm a bit confused on how to best do it. I know Householder factorization, but I'm confused on how to do it.

So $Q = I_4 - 2 \times \frac{1}{25} \left( \begin{array}{ccc} -2\\ 0 \\ 4 \\ 0 \end{array} \right) \times (-2,0,4,0)$.

Then, do you just do $QA = R$. I was wondering is there an easier way to work out the QR factorization of this.

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1 Answer 1

up vote 2 down vote accepted

(Note: your calculation contains an error -- the fraction in your expression for $Q$ should be $\|(-2,0,4,0)\|^2=\frac1{20}$ rather than $\|(3,0,4,0)\|^2=\frac1{25}$.)

If by "QR factorization" you mean the QR factorization algorithm, I don't think there is an easier way to decompose $A$. After all, you have to follow the steps in the algorithm. However, if you simply want to decompose $A$ into the form of $QR$, there is a much easier way: note that the two columns of $A$ are already orthogonal to each other. So, if you normalize them and complete the normalized $A$ to an orthogonal matrix $Q$, then $A=QD$ for some diagonal (hence triangular) $D$. More specifically, let $$ Q=\frac{1}{5}\begin{pmatrix} 3&0&4&0\\ 0&3&0&4\\ 4&0&-3&0\\ 0&4&0&-3\\ \end{pmatrix}. $$ (Note that the first two columns of $Q$ are multiples of the first two columns of $A$.) Then $$ A=Q\begin{pmatrix} 5&0\\ 0&5\\ 0&0\\ 0&0 \end{pmatrix}. $$

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Thanks for that. I didn't see that trick. Is there a more formal way to work out that $Q$. As I was wondering if you can do the trick for any matrix with linearly independent columns. –  simplicity Dec 3 '12 at 3:18
    
I don't think this trick has a wider application. QR factorization involves orthogonality, which is a stronger requirement than linear independence. Actually our current problem is easy not just because the columns of $A$ are orthogonal, but also because $A$ is comprised blocks of multiples of $I_2$, so that we can immediately complete its columns to an orthogonal basis without any hand calculations. Should $A$ contains other orthogonal columns, the difficulty of decomposing it into $QR$ would be higher. –  user1551 Dec 3 '12 at 12:15

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