QR decomposition

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In linear algebra, the QR decomposition of a matrix is a decomposition of the matrix into an orthogonal and a 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.

1 Definition
2 Computing the QR decomposition
3 Computing QR by means of Gram-Schmidt
- 3.1 Example
4 Computing QR by means of Householder reflections
- 4.1 Example
5 Computing QR by means of Givens rotations
- 5.1 Example
6 See also
7 External links

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Definition

A QR decomposition of a real square matrix A is a decomposition of A as

where Q is an orthogonal matrix (meaning that Q^TQ = I) and R is an upper triangular matrix.

More generally, we can factor an <math>m</math>×<math>n</math> matrix (with m ≥ n) of full rank as the product of an <math> m</math>×<math>n</math> unitary matrix (in the sense that Q^∗Q = I) and an <math>n</math>×<math> n</math> upper triangular matrix.

This factorization is unique if we require that the diagonal elements of R are positive.

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Computing the QR decomposition

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.

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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 a_is 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 e_ks), 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>.

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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^{-16}  &                     175           &            -70 \\
   -1.77635683940025 \times 10^{-15}  &  -5.32907051820075 \times 10^{-14}           &             35

\end{pmatrix}. </math>

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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 <math>\mathbf{x}</math> be an arbitrary m-dimensional column vector such that ||<math>\mathbf{x}</math>|| = |α| for a scalar α (if the algorithm is implemented using floating-point arithmetic, then α should get the same sign as the first coordinate of <math>\mathbf{x}</math> to avoid loss of significance).

Then, where <math>\mathbf{e}_1</math> 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 row).

<math>Q_1A = \begin{bmatrix}

                  \alpha_1&\star&\dots&\star\\
                     0    &     &     &    \\
                  \vdots  &     &  A' &    \\
                     0    &     &     & \end{bmatrix}</math>

This can be repeated for A′ (obtained from A by deleting the first row and first column), resulting in a Householder 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 the Gram-Schmidt method above.

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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 find a reflection that transforms the first column of matrix A, 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} = \mathbf{x} - \alpha\mathbf{e}_1,</math>

and

<math>\mathbf{v} = {\mathbf{u}\over\|\mathbf{u}\|},</math>.

Here,

<math>\alpha = 14</math> and <math>\mathbf{x} = \mathbf{a}_1 = (12, 6, -4)^T</math>

Therefore

<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>

Now observe:

<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

<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.

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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 Givens 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.

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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 = \arctan({-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>.

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