Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$ A=Q\begin{pmatrix} 5&0\\ 0&5\\ 0&0\\ 0&0 \end{pmatrix}. $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.