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Computing QR factorizations

To compute an LU factorization, we follow elimination rules to introduce zeros into the lower triangle of the matrix, leaving only the $\mathbf{U}$ factor. The row operations are themselves triangular and can be combined into the $\mathbf{L}$ factor. For the QR factorization, the game is again to introduce zeros into the lower triangle, but the rules have changed; now the row operations must be done orthogonally. Thanks to the orthogonal matrix theorem, the product of orthogonal operations will also be orthogonal, providing us with the final $\mathbf{Q}$ of the QR.

Householder reflections

A Householder reflector is a particular type of orthogonal matrix $\mathbf{P}$. The reflection is customized for a particular given vector $\mathbf{z}$ so that $\mathbf{P}\mathbf{z}$ is nonzero only in the first element. Since orthogonal matrices preserve the 2-norm, we must have

(71) $$\begin{split}\mathbf{P}\mathbf{z} = \begin{bmatrix} \pm \| \mathbf{z} \|\\0 \\ \vdots \\ 0 \end{bmatrix} = \pm \| \mathbf{z} \| \mathbf{e}_1.\end{split}$$

(Recall that $\mathbf{e}_k$ is the $k$th column of the identity matrix.) We choose the positive sign for our discussion, but see qrfact and an exercise for important computational details.

Given $\mathbf{z}$, let

(72) $$\mathbf{v} = \| \mathbf{z} \| \mathbf{e}_1-\mathbf{z}.$$

Then the reflector $\mathbf{P}$ is defined by

(73) $$\mathbf{P} = \mathbf{I} - 2\frac{\mathbf{v} \mathbf{v}^T}{\mathbf{v}^T\mathbf{v}}$$

if $\mathbf{v}\neq\boldsymbol{0}$, or $\mathbf{P}=\mathbf{I}$ if $\mathbf{v}=\boldsymbol{0}$. Note that $\mathbf{v}^T\mathbf{v}$ is a scalar and can appear in a denominator, while the outer product $\mathbf{v} \mathbf{v}^T$ is $n\times n$. It is straightforward to show that $\mathbf{P}$ has the following key properties.

Theorem 29 ((Householder reflector))

Let $\mathbf{v}=\| \mathbf{z} \|\mathbf{e}_1-\mathbf{z}$ and let $\mathbf{P}$ be given by (73). Then $\mathbf{P}$ is symmetric and orthogonal, and $\mathbf{P}\mathbf{z}=\| \mathbf{z} \|\mathbf{e}_1$.

Proof

The case $\mathbf{v}=\boldsymbol{0}$ is obvious. For $\mathbf{v}\neq\boldsymbol{0}$, the proofs of symmetry and orthogonality are left to the exercises. As for the last fact, we simply compute

(74) $$\mathbf{P}\mathbf{z} = \mathbf{z} - 2 \frac{\mathbf{v} \mathbf{v}^T \mathbf{z}}{\mathbf{v}^T\mathbf{v}} = \mathbf{z} - 2 \frac{\mathbf{v}^T \mathbf{z}}{\mathbf{v}^T\mathbf{v}} \mathbf{v},$$

and, since $\mathbf{e}_1^T\mathbf{z}=z_1$,

$$\begin{split}\begin{split} \mathbf{v}^T\mathbf{v} &= \| \mathbf{z} \|^2 - 2 \| \mathbf{z} \| z_1 + \mathbf{z}^T\mathbf{z} = 2\| \mathbf{z} \|(\| \mathbf{z} \|-z_1),\\ \mathbf{v}^T\mathbf{z} &= \| \mathbf{z} \|z_1 - \mathbf{z}^T\mathbf{z} = -\| \mathbf{z} \|\bigl(\| \mathbf{z} \|-z_1\bigr), \end{split}\end{split}$$

leading finally to

$$\mathbf{P}\mathbf{z} = \mathbf{z} - 2\cdot \frac{-\| \mathbf{z} \| \bigl(\| \mathbf{z} \|-z_1\bigr)}{2\| \mathbf{z} \| \bigl(\| \mathbf{z} \|-z_1\bigr)} \mathbf{v} = \mathbf{z} + \mathbf{v} = \| \mathbf{z} \|\mathbf{e}_1.$$

The reason $\mathbf{P}$ is called a reflector is sketched in Fig. 4.

Fig. 4 Householder reflector.

The vector $\mathbf{v}$ defines an $n-1$ dimensional subspace $S$ perpendicular to it. Elementary vector analysis shows that $\mathbf{v}(\mathbf{v}^T\mathbf{z})/(\mathbf{v}^T\mathbf{v})$ is the vector projection of $\mathbf{z}$ along $\mathbf{v}$. By subtracting this projection from $\mathbf{z}$, we end up in $S$, but by subtracting twice the projection we get a reflection through $S$. This reflection occurs when $\mathbf{P}$ is applied to any vector, but when it is applied to $\mathbf{z}$, the result ends up on the $x_1$-axis.

Factorization algorithm

The QR factorization is computed by using successive Householder reflections to introduce zeros in one column at a time.

Demo

Householder QR

The QR factorization is computed by using successive Householder reflections to introduce zeros in one column at a time. We first show the process for a small numerical example in Householder QR.

You may be wondering what happened to the $\mathbf{Q}$ in Householder QR. Each Householder reflector is orthogonal but not full-size. We have to pad it out to represent algebraically the fact that a block of the first rows are left alone. Given a reflector $\mathbf{P}_k$ that is of square size $m-k+1$, we define

$$\begin{split}\mathbf{Q}_k = \begin{bmatrix} \mathbf{I}_{k-1} & \boldsymbol{0} \\ \boldsymbol{0} & \mathbf{P}_k \end{bmatrix}.\end{split}$$

It is trivial to show that $\mathbf{Q}_k$ is also orthogonal. Then

$$\mathbf{Q}_n \mathbf{Q}_{n-1}\cdots \mathbf{Q}_1 \mathbf{A} = \mathbf{R}.$$

But $\mathbf{Q}_n \mathbf{Q}_{n-1}\cdots \mathbf{Q}_1$ is orthogonal too, and we multiply on the left by its transpose to get $\mathbf{A}=\mathbf{Q}\mathbf{R}$, where $\mathbf{Q} = (\mathbf{Q}_n \mathbf{Q}_{n-1}\cdots \mathbf{Q}_1)^T$. We don’t even need to form these matrices explicitly. Writing

$$\mathbf{Q}^T = \mathbf{Q}_n \mathbf{Q}_{n-1}\cdots \mathbf{Q}_1 = \mathbf{Q}_n \Bigl( \mathbf{Q}_{n-1}\bigl(\cdots (\mathbf{Q}_1\mathbf{I})\cdots\bigr)\Bigr),$$

we can build $\mathbf{Q}^T$ iteratively by starting with the identity and doing the same row operations as on $\mathbf{A}$. Creating $\mathbf{Q}^T$ with row operations on $\mathbf{I}$ uses much less memory than building the $\mathbf{Q}_k$ matrices explicitly.

The algorithm we have described is encapsulated in qrfact. There is one more refinement in it, though. As indicated by (74), the application of a reflector $\mathbf{P}$ to a vector does not require the formation of the matrix $\mathbf{P}$ itself. Its effect can be computed directly from the vector $\mathbf{v}$, as is shown in qrfact.

Function 30 (qrfact)

QR factorization by Householder reflections.

"""
qrfact(A)

QR factorization by Householder reflections. Returns Q and R.
"""
function qrfact(A)

    m,n = size(A)
    Qt = Matrix(Diagonal(ones(m)))
    R = float(copy(A))
    for k in 1:n
      z = R[k:m,k]
      v = [ -sign(z[1])*norm(z) - z[1]; -z[2:end] ]
      nrmv = norm(v)
      if nrmv < eps() continue; end  # skip this iteration
      v = v / nrmv;                  # simplifies other formulas
      # Apply the reflection to each relevant column of A and Q
      for j in 1:n
        R[k:m,j] -= v*( 2*(v'*R[k:m,j]) )
      end
      for j in 1:m
        Qt[k:m,j] -= v*( 2*(v'*Qt[k:m,j]) )
      end
    end
    
    return Qt',triu(R)
end

The Julia command qr works similarly to, but more efficiently than, qrfact. It finds the factorization in $\sim(2m n^2-n^3/3)$ flops asymptotically.

Exercises

1. ⌨ Find a Householder reflector $\mathbf{P}$ such that

   $$\begin{split}\mathbf{P} \begin{bmatrix} -6 \\ 2 \\ 9 \end{bmatrix} = \begin{bmatrix} 11\\0\\0 \end{bmatrix}.\end{split}$$

2. ✍ Prove the unfinished items in the reflector theorem, namely that a Householder reflector $\mathbf{P}$ is symmetric and orthogonal.

3. ✍ Let $\mathbf{P}$ be a Householder reflector as in (73).

   (a) Find a vector $\mathbf{u}$ such that $\mathbf{P}\mathbf{u} = -\mathbf{u}$. (Fig. 4 may be of help.)

   (b) What algebraic condition is necessary and sufficient for a vector $\mathbf{w}$ to satisfy $\mathbf{P}\mathbf{w}=\mathbf{w}$? In $n$ dimensions, how many such linearly independent vectors are there?

4. ✍ Under certain circumstances, computing the vector $\mathbf{v}$ in (72) could lead to subtractive cancellation, which is why line 13 of qrfact reads as it does. Devise an example that causes subtractive cancellation if (72) is used.

5. ✍ Suppose QR factorization is used to compute the solution of a square linear system, $\mathbf{A}\mathbf{x}=\mathbf{b}$; i.~e., let $m=n$.

   (a) Find an asymptotic flop count for this procedure, and compare to the LU factorization algorithm.

   (b) Show that $\kappa_2(\mathbf{A})=\kappa_2(\mathbf{R})$.

6. ✍ Show that $\kappa_2(\mathbf{A})=\kappa_2(\mathbf{R})$ when $\mathbf{A}$ is not square. (Hint: You can’t take an inverse of $\mathbf{A}$ or $\mathbf{R}$.)

7. ⌨ Modify qrfact so that the loops in lines 18–23 are removed. In other words, update the matrices A and Q with a single statement for each. Test your code against the original code on an example to verify it.

8. Another algorithmic technique for orthogonally introducing zeros into a matrix is the Givens rotation. Given a 2-vector $[\alpha\, \beta]$, it defines an angle $\theta$ such that

   $$\begin{split}\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta^2 \\ 0 \end{bmatrix}.\end{split}$$

   (a) ✍ Given $\alpha$ and $\beta$, show how to compute $\theta$.

   (b) ⌨ Given the vector $\mathbf{z}=[1\;2\;3\;4\;5]^T$, use Julia to find a sequence of Givens rotations that transforms $\mathbf{z}$ into the vector $\| \mathbf{z} \|\mathbf{e}_1$. (Hint: You can operate only on pairs of elements at a time, introducing a zero at the lower of the two positions.)

The QR factorization Next steps