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I have a hard time finding the operation count of QR factorization when using Householder transformations. The answer is $2mn^2 - \frac{2n^3}{3}$, but have no clue on how to get this count following some procedure.

Could anyone help me go through this step by step?

Stefan4024
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onimoni
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    Did you try to go through the algorithm, writing down the operation s per iteration? It might help. – onimoni Sep 22 '13 at 10:58

1 Answers1

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Suppose you have a $m \times n$ matrix $A$, $m \geq n$, and you want to perform QR factorization using Householder transformations.

You will have to go through $A$ as follows:

  1. Start with the whole matrix

  2. Proceed with the modified matrix of original size without it's first row and first column.

  3. Proceed with second modified matrix without first two rows and columns.

  4. And so on...

In a more compact way this is stated as $A_{k:m,k:n}$ for every iteration step $k$. Now the (part of the) algorithm (which does most of the work) can be denoted similarly, i.e. per iteration you have $A_{k:m,k:n} = A_{k:m,k:n} - 2v_k (v_{k}^TA_{k:m,k:n})$.

Now the operation you want to count are in these iterations. Per iteration you have:

  1. $(m-k)(n-k)$ for the outer product $2v_k(\dots)$
  2. $2(m-k)(n-k)$ for the products $v_{k}^TA_{k:m,k:n}$
  3. $(m-k)(n-k)$ for the subtraction $A_{k:m,k:n} - \dots$

So that is $4(m-k)(n-k)$ in total. Which you do for every iteration (!), so you get

$\sum\limits_{k=1}^{n} 4(m-k)(n-k) =4 \sum\limits_{k=1}^{n} (mn-k(m+n)+k^2)$

$\approx 4mn^2 - 4(m+n)n^2/2 + 4n^3/3 = 2mn^2-2n^3/3$

Nachi
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onimoni
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