QR decomposition
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In linear algebra, the QR decomposition of a matrix <math>A<math> is a factorization expressing <math>A<math> as
- <math>A = QR<math>
where <math>Q<math> is an orthogonal matrix (<math>QQ^T = I<math>), and <math>R<math> is an upper triangular matrix.
The QR decomposition is often used to solve the linear least squares problem. The QR decomposition is also the basis for a particular eigenvalue algorithm, the QR algorithm.
There are several methods for actually computing the QR decomposition, such as by means of Givens rotations, Householder transformations, or the Gram-Schmidt decomposition. Each has a number of advantages and disadvantages. (The matrices Q and R are not uniquely determined, so different methods may produce different decompositions.)
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Computing QR by means of Gram-Schmidt
Recall the Gram-Schmidt method, with the vectors to be considered in the process as columns of the matrix <math>A=(\mathbf{a}_1| \cdots|\mathbf{a}_n)<math>. Then
- <math>
\mathbf{u}_1 = \mathbf{a}_1, \qquad\mathbf{e}_1 = {\mathbf{u}_1 \over ||\mathbf{u}_1||}<math>
- <math>
\mathbf{u}_2 = \mathbf{a}_2-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2, \qquad\mathbf{e}_2 = {\mathbf{u}_2 \over ||\mathbf{u}_2||}<math>
- <math>
\mathbf{u}_3 = \mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3, \qquad\mathbf{e}_3 = {\mathbf{u}_3 \over ||\mathbf{u}_3||} <math>
- <math>\vdots<math>
- <math>
\mathbf{u}_k = \mathbf{a}_k-\sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k,\qquad\mathbf{e}_k = {\mathbf{u}_k\over||\mathbf{u}_k||} <math>
Naturally then, we rearrange the equations so the ais are the subject, to get the following
- <math>\mathbf{a}_1 = \mathbf{e}_1||\mathbf{u}_1|| <math>
- <math>\mathbf{a}_2 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2+\mathbf{e}_2||\mathbf{u}_2|| <math>
- <math>\mathbf{a}_3 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3+\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3+\mathbf{e}_3||\mathbf{u}_3|| <math>
- <math>\vdots<math>
- <math>\mathbf{a}_k = \sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k+\mathbf{e}_k||\mathbf{u}_k|| <math>
Each of these projections of the vectors <math>\mathbf{a}_i<math> onto one of these <math>e_j<math> are merely the inner product of the two, since the vectors are normed.
Now these equations can be written in matrix form, viz.,
- <math>\left(\mathbf{e}_1\left|\ldots\right|\mathbf{e}_n\right)
\begin{pmatrix} ||\mathbf{u}_1|| & \langle\mathbf{e}_1,\mathbf{a}_2\rangle & \langle\mathbf{e}_1,\mathbf{a}_3\rangle & \ldots \\ 0 & ||\mathbf{u}_2|| & \langle\mathbf{e}_2,\mathbf{a}_3\rangle & \ldots \\ 0 & 0 & ||\mathbf{u}_3|| & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}<math> But the product of each row and column of the matrices above give us a respective column of A that we started with, and together, they give us the matrix A, so we have factorized A into an orthogonal matrix Q (the matrix of eks), via Gram Schmidt, and the obvious upper triangular matrix as a remainder R.
Alternatively, <math>\begin{matrix} R \end{matrix}<math> can be calculated as follows:
Recall that <math> \begin{matrix}Q\end{matrix} = \left(\mathbf{e}_1\left|\ldots\right|\mathbf{e}_n\right). <math> Then, we have
- <math>
\begin{matrix} R = Q^{T}A = \end{matrix} \begin{pmatrix} \langle\mathbf{e}_1,\mathbf{a}_1\rangle & \langle\mathbf{e}_1,\mathbf{a}_2\rangle & \langle\mathbf{e}_1,\mathbf{a}_3\rangle & \ldots \\ 0 & \langle\mathbf{e}_2,\mathbf{a}_2\rangle & \langle\mathbf{e}_2,\mathbf{a}_3\rangle & \ldots \\ 0 & 0 & \langle\mathbf{e}_3,\mathbf{a}_3\rangle & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}. <math> Note that <math>\langle\mathbf{e}_j,\mathbf{a}_j\rangle = ||\mathbf{u}_j||,<math> <math>\langle\mathbf{e}_j,\mathbf{a}_k\rangle = 0 \mathrm{~~for~~} j > k,<math> and <math> QQ^{T} = I <math>, so <math> Q^{T} = Q^{-1} <math>.
Example
Consider the decomposition of
- <math>A =
\begin{pmatrix} 12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix} .<math>
Recall the orthogonal matrix <math>Q<math> such that
- <math>
\begin{matrix}
Q\,Q^{T} = I.
\end{matrix} <math>
Then, we can calculate <math>Q<math> by means of Gram-Schmidt as follows:
- <math>
U = \begin{pmatrix} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{pmatrix} = \begin{pmatrix} 12 & -69 & -58 \\ 6 & 158 & 6 \\ -4 & 30 & -165 \end{pmatrix}; <math>
- <math>
Q = \begin{pmatrix} \frac{\mathbf u_1}{||\mathbf u_1||} & \frac{\mathbf u_2}{||\mathbf u_2||} & \frac{\mathbf u_3}{||\mathbf u_3||} \end{pmatrix} = \begin{pmatrix}
6/7 & -69/175 & -58/175 \\
3/7 & 158/175 & 6/175 \\
-2/7 & 6/35 & -33/35
\end{pmatrix}; <math>
Thus, we have
- <math>
\begin{matrix}
A = Q\,Q^{T}A = Q R;
\end{matrix} <math>
- <math>
\begin{matrix}
R = Q^{T}A =
\end{matrix} \begin{pmatrix}
14 & 21 & -14 \\
0 & 175 & -70 \\
0 & 0 & 35
\end{pmatrix}. <math>
Considering numerical errors of finite precision operation in MATLAB, we have that
- <math>
\begin{matrix}
Q =
\end{matrix} \begin{pmatrix}
0.857142857142857 & -0.394285714285714 & -0.331428571428571 \\
0.428571428571429 & 0.902857142857143 & 0.034285714285714 \\
-0.285714285714286 & 0.171428571428571 & -0.942857142857143
\end{pmatrix}; <math>
- <math>
\begin{matrix}
R =
\end{matrix} \begin{pmatrix}
14 & 21 & -14 \\
1.11022302462516 \times 10^{-016} & 175 & -70 \\
-1.77635683940025 \times 10^{-015} & -5.32907051820075 \times 10^{-014} & 35
\end{pmatrix}. <math>
Computing QR by means of Householder reflections
A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane. We can use this property to calculate the QR factorization of a matrix.
Q can be used to reflect a vector in such a way that all coordinates but one disappear. Let x be an arbitrary m-dimensional column vector of length |α| (for numerical reasons α should get the same sign as the first coordinate of x).
Then, where e1 is the vector (1,0,...,0)T, and || || the euclidean norm, set
- <math>\mathbf{u} = \mathbf{x} - \alpha\mathbf{e}_1,<math>
- <math>\mathbf{v} = {\mathbf{u}\over||\mathbf{u}||},<math>
- <math>Q = I - 2 \mathbf{v}\mathbf{v}^T.<math>.
<math>Q<math> is a Householder matrix and
- <math>Qx = (\alpha\ , 0, \cdots, 0)^T<math>.
This can be used to gradually transform an m-by-n matrix A to upper triangular form. First, we multiply A with the Householder matrix Q1 we obtain when we choose the first matrix column for x. This results in a matrix QA with zeros in the left column (except for the first line).
- <math>Q_1A = \begin{bmatrix}
\alpha_1&\star&\dots&\star\\
0 & & & \\
\vdots & & A' & \\
0 & & & \end{bmatrix}<math>
This can be repeated for A' resulting in a Housholder matrix Q'2. Note that Q'2 is smaller than Q1. Since we want it really to operate on Q1A instead of A' we need to expand it to the upper left, filling in a 1, or in general:
- <math>Q_k = \begin{pmatrix}
I_{k-1} & 0\\
0 & Q_k'\end{pmatrix}<math>
After <math>t<math> iterations of this process, <math>t = min(m - 1, n)<math>,
- <math> R = Q_t \cdots Q_2Q_1A<math>
is a upper triangular matrix. So, with
- <math> Q = Q_1Q_2 \cdots Q_t<math>
<math>A = QR<math> is a QR decomposition of <math>A<math>.
This method has greater numerical stability than using the Gram-Schmidt method above. .
Example
Let us calculate the decomposition of
- <math>A = \begin{pmatrix}
12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix}<math>
We need to find a reflection that takes the vector <math>\mathbf{a}_1 = (12, 6, -4)^T<math> to <math>\| \;\mathbf{a}_1\| \;\mathrm{e}_1 = (14, 0, 0)^T <math>.
Now, <math>\mathbf{u} = (-2, 6, -4)^T<math> and <math>\mathbf{v} = {1 \over \sqrt{14}}(-1, 3, -2)^T<math>, and then
- <math>Q_1 = I - {2 \over \sqrt{14}\sqrt{14}} \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix}\begin{pmatrix} -1 & 3 & -2 \end{pmatrix}<math>
- <math> = I - {1 \over 7}\begin{pmatrix}
1 & -3 & 2 \\ -3 & 9 & -6 \\ 2 & -6 & 4 \end{pmatrix}<math>
- <math> = \begin{pmatrix}
6/7 & 3/7 & -2/7 \\ 3/7 &-2/7 & 6/7 \\ -2/7 & 6/7 & 3/7 \\ \end{pmatrix}<math>
Observe now:
- <math>Q_1A = \begin{pmatrix}
14 & 21 & -14 \\ 0 & -49 & -14 \\ 0 & 168 & -77 \end{pmatrix}<math> So we already have almost a triangular matrix. We only need to zero the (3, 2) entry.
Take the (1, 1) minor, and then apply the process again to
- <math>A' = M_{11} = \begin{pmatrix}
-49 & -14 \\ 168 & -77 \end{pmatrix}<math> By the same method as above, we obtain the matrix of the Householder transformation to be,
- <math>Q_2 = \begin{pmatrix}
1 & 0 & 0 \\ 0 & -7/25 & 24/25 \\ 0 & 24/25 & 7/25 \end{pmatrix}<math> after performing a direct sum with 1 to make sure the next step in the process works properly.
Now, we find
- <math>Q=Q_1Q_2=\begin{pmatrix}
6/7 & -69/175 & 58/175 \\ 3/7 & 158/175 & -6/175 \\ -2/7 & 6/35 & 33/35 \end{pmatrix} <math>
- <math>R=Q^\top A=\begin{pmatrix}
14 & 21 & -14 \\ 0 & 175 & -70 \\ 0 & 0 & -35 \end{pmatrix}<math> The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition.
Computing QR by means of Givens rotations
QR decompositions can also be computed with a series of Givens rotations. Each rotation zeros an element in the subdiagonal of the matrix, forming the R matrix. The concatenation of all the Given rotations forms the orthogonal Q matrix.
In practice, Givens rotations are not actually performed by building a whole matrix and doing a matrix multiplication. A Givens rotation procedure is used instead which does the equivalent of the sparse Givens matrix multiplication, without the extra work of handling the sparse elements. The Givens rotation procedure is useful in situations where only a relatively few off diagonal elements need to be zeroed, and is more easily parallelized than Householder transformations.
Example
Let us calculate the decomposition of
- <math>A = \begin{pmatrix}
12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix}<math>
First, we need to form a rotation matrix that will zero the lowermost left element, <math>\mathbf{a}_{31} = -4<math>. We form this matrix using the Givens rotation method, and call the matrix <math>G_1<math>. We will first rotate the vector <math>(6,-4)<math>, to point along the X axis. This vector has an angle <math>\theta = tan^{-1}({-4 \over 6})<math>. We create the orthogonal Givens rotation matrix, <math>G_1<math>:
- <math>G_1 = \begin{pmatrix}
1 & 0 & 0 \\ 0 & cos(\theta) & sin(\theta) \\ 0 & -sin(\theta) & cos(\theta) \end{pmatrix}<math>
- <math>\approx \begin{pmatrix}
1 & 0 & 0 \\ 0 & 0.83205 & -0.55470 \\ 0 & 0.55470 & 0.83205 \end{pmatrix}<math>
And the result of <math>G_1A<math> now has a zero in the <math>\mathbf{a}_{31}<math> element.
- <math>G_1A \approx \begin{pmatrix}
12 & -51 & 4 \\ 7.21110 & 125.63959 & -33.83671 \\ 0 & 112.60414 & -71.83368 \end{pmatrix}<math>
We can similarly form Givens matrices <math>G_2<math> and <math>G_3<math>, which will zero the sub-diagonal elements <math>a_{21}<math> and <math>a_{32}<math>, forming a rectangular matrix <math>R<math>. The orthogonal matrix <math>Q^T<math> is formed from the concatenation of all the Givens matrices <math>Q^T = G_3G_2G_1<math>. Thus, we have <math> G_3G_2G_1A= Q^TA = R<math>, and the QR decomposition is <math>A = QR<math>.de:QR-Zerlegung fr:Décomposition QR ja:QR分解