A rectangular, $A \in \mathbb{R}^{m \times n}$ matrix, where $m \ge n$, can be decomposed (QR factorization): $$A = \begin{bmatrix}Q_1 | Q_2 \end{bmatrix}\begin{bmatrix}R\\0\end{bmatrix}$$ where $Q_1$ and $Q_2$ has orthonormal columns, and $R$ is upper triangular.

I'm implementing a routine (based on Householder reflections) which calculates $Q_1$ and $R$ (so called thin/reduced QR decomposition).

My question is: is it possible to calculate $Q_1$ without calculating $Q_2$? The problem is that a Householder matrix is $\mathbb{R}^{m \times m}$, and $Q_1 \in \mathbb{R}^{m \times n}$, so I cannot multiply them. My routine currently calculates $Q=[Q_1|Q_2]$, and then throws away the $Q_2$ part.

  • $\begingroup$ $m \le n$? Would Gram-Schmidt suit what you need? (E.g. here) $\endgroup$ Jan 27, 2019 at 11:47
  • 1
    $\begingroup$ @PaulAljabar: in the case of thin QR decomposition, $A$ has more rows than columns (for the other case, "wide'' matrices, thin QR factorization is not a thing). I'd like to keep using Householder reflection, as it is numerically more stable. $\endgroup$
    – geza
    Jan 27, 2019 at 11:57

1 Answer 1


Yes, it is possible, and it is quite straightforward:$$R=H_nH_{n-1}\dots H_1A,$$ where $H_*$ are the Householder reflectors. So, $$Q = H_1^T\dots H_{n-1}^TH_n^T.$$

As $H_*$ are symmetric: $$Q = H_1\dots H_{n-1}H_n.$$

Now, to compute $Q_1$, we have:$$Q_1 = H_1\dots H_{n-1}H_nI^{m \times n},$$where $I^{m \times n} \in \mathrm{R}^{m \times n}$ is the rectangular identity matrix.

$Q_1$ can be calculated backwards: first calculate $H_nI^{m \times n}$, the result is a $m \times n$ matrix. Then multiply left this with $H_{n-1}$, then $H_{n-2}$, and so on.

This way, the full $Q$ doesn't have to be formulated.


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