QR decomposition

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In linear algebra, the QR decomposition (also called the QR factorization) of a matrix is a decomposition of the matrix into an orthogonal and a right 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 Connection to a determinant or a product of eigenvalues
4 See also
5 References
6 External links

[edit] Definition

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

$A = QR, \,$

where Q is an orthogonal matrix (meaning that Q^TQ = I ) and R is an upper triangular matrix (also called right triangular matrix). Analogously, we can define the QL, RQ, and LQ decompositions of A (with L being a lower triangular matrix in this case).

More generally, we can factor a complex $m$ × $n$ matrix (with m ≥ n) as the product of an $m$ × $m$ unitary matrix and an $m$ × $n$ upper triangular matrix. An alternative definition is decomposing a complex $m$ × $n$ matrix (with m ≥ n) as the product of an $m$ × $n$ matrix with orthogonal columns and an $n$ × $n$ upper triangular matrix; Golub & Van Loan (1996, §5.2) call this the thin QR factorization.

If A is nonsingular, then this factorization is unique if we require that the diagonal elements of R are positive.

[edit] Computing the QR decomposition

There are several methods for actually computing the QR decomposition, such as by means of the Gram–Schmidt process, Householder transformations, or Givens rotations. Each has a number of advantages and disadvantages.

[edit] Using the Gram-Schmidt process

For more details on this topic, see Gram-Schmidt#Numerical stability.

Consider the Gram–Schmidt process, with the vectors to be considered in the process as the columns of the matrix $A=(\mathbf{a}_1| \cdots|\mathbf{a}_n)$ . We define $\mathrm{proj}_{\mathbf{e}}\mathbf{a} = \frac{\left\langle\mathbf{e},\mathbf{a}\right\rangle}{\left\langle\mathbf{e},\mathbf{e}\right\rangle}\mathbf{e}$ where $\left\langle\mathbf{v},\mathbf{w}\right\rangle =\mathbf{v}^T\mathbf{w}$ .

Then

$\mathbf{u}_1 = \mathbf{a}_1, \qquad\mathbf{e}_1 = {\mathbf{u}_1 \over \|\mathbf{u}_1\|}$

$\mathbf{u}_2 = \mathbf{a}_2-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2, \qquad\mathbf{e}_2 = {\mathbf{u}_2 \over \|\mathbf{u}_2\|}$

$\mathbf{u}_3 = \mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3, \qquad\mathbf{e}_3 = {\mathbf{u}_3 \over \|\mathbf{u}_3\|}$

$\vdots$

$\mathbf{u}_k = \mathbf{a}_k-\sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k,\qquad\mathbf{e}_k = {\mathbf{u}_k\over\|\mathbf{u}_k\|}$

We then rearrange the equations above so that the $\mathbf{a}_i$ s are on the left, producing the following equations.

$\mathbf{a}_1 = \mathbf{e}_1\|\mathbf{u}_1\|$

$\mathbf{a}_2 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2+\mathbf{e}_2\|\mathbf{u}_2\|$

$\mathbf{a}_3 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3+\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3+\mathbf{e}_3\|\mathbf{u}_3\|$

$\vdots$

$\mathbf{a}_k = \sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k+\mathbf{e}_k\|\mathbf{u}_k\|$

Note that since the $\mathbf{e}_i$ are unit vectors, we have the following.

$\mathbf{a}_1 = \mathbf{e}_1\|\mathbf{u}_1\|$

$\mathbf{a}_2 = \left\langle\mathbf{e}_1,\mathbf{a}_2\right\rangle\mathbf{e}_1 +\mathbf{e}_2\|\mathbf{u}_2\|$

$\mathbf{a}_3 = \left\langle\mathbf{e}_1,\mathbf{a}_3\right\rangle\mathbf{e}_1 +\left\langle\mathbf{e_2},\mathbf{a}_3\right\rangle\mathbf{e}_2 +\mathbf{e}_3\|\mathbf{u}_3\|$

$\vdots$

$\mathbf{a}_k = \sum_{j=1}^{k-1}\left\langle\mathbf{e}_j,\mathbf{a}_k\right\rangle\mathbf{e}_j +\mathbf{e}_k\|\mathbf{u}_k\|$

Now the right sides of these equations can be written in matrix form as follows:

$\left(\mathbf{e}_1\left|\ldots\right|\mathbf{e}_n\right) \begin{pmatrix} \|\mathbf{u}_1\| & \langle\mathbf{e}_1,\mathbf{a}_2\rangle & \langle\mathbf{e}_1,\mathbf{a}_3\rangle & \ldots \\ 0 & \|\mathbf{u}_2\| & \langle\mathbf{e}_2,\mathbf{a}_3\rangle & \ldots \\ 0 & 0 & \|\mathbf{u}_3\| & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$

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, $\begin{matrix} R \end{matrix}$ can be calculated as follows:

Recall that $\begin{matrix}Q\end{matrix} = \left(\mathbf{e}_1\left|\ldots\right|\mathbf{e}_n\right).$ Then, we have

$\begin{matrix} R = Q^{T}A = \end{matrix} \begin{pmatrix} \langle\mathbf{e}_1,\mathbf{a}_1\rangle & \langle\mathbf{e}_1,\mathbf{a}_2\rangle & \langle\mathbf{e}_1,\mathbf{a}_3\rangle & \ldots \\ 0 & \langle\mathbf{e}_2,\mathbf{a}_2\rangle & \langle\mathbf{e}_2,\mathbf{a}_3\rangle & \ldots \\ 0 & 0 & \langle\mathbf{e}_3,\mathbf{a}_3\rangle & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$

Note that $\langle\mathbf{e}_j,\mathbf{a}_j\rangle = \|\mathbf{u}_j\|,$ $\langle\mathbf{e}_j,\mathbf{a}_k\rangle = 0 \mathrm{~~for~~} j > k,$ and $Q Q T = I$ , so $Q T = Q - 1$ .

[edit] Example

Consider the decomposition of

$A = \begin{pmatrix} 12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix} .$

Recall that an orthogonal matrix $Q$ has the property

$\begin{matrix} Q\,Q^{T} = I. \end{matrix}$

Then, we can calculate $Q$ by means of Gram-Schmidt as follows:

$U = \begin{pmatrix} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{pmatrix} = \begin{pmatrix} 12 & -69 & -58/5 \\ 6 & 158 & 6/5 \\ -4 & 30 & -33 \end{pmatrix};$

$Q = \begin{pmatrix} \frac{\mathbf u_1}{\|\mathbf u_1\|} & \frac{\mathbf u_2}{\|\mathbf u_2\|} & \frac{\mathbf u_3}{\|\mathbf u_3\|} \end{pmatrix} = \begin{pmatrix} 6/7 & -69/175 & -58/175 \\ 3/7 & 158/175 & 6/175 \\ -2/7 & 6/35 & -33/35 \end{pmatrix};$

Thus, we have

$\begin{matrix} A = Q\,Q^{T}A = Q R; \end{matrix}$

$\begin{matrix} R = Q^{T}A = \end{matrix} \begin{pmatrix} 14 & 21 & -14 \\ 0 & 175 & -70 \\ 0 & 0 & 35 \end{pmatrix}.$

[edit] Relation to RQ decomposition

The RQ decomposition transforms a matrix A into the product of an upper triangular matrix R (also known as right-triangular) and an orthogonal matrix Q. The only difference from QR decomposition is the order of these matrices.

QR decomposition is Gram-Schmidt orthogonalization of columns of A, started from the first column.

RQ decomposition is Gram-Schmidt orthogonalization of rows of A, started from the last row.

[edit] Application to computer vision

The RQ decomposition can be used directly to extract the intrinsic camera parameters from the projection matrix P.

[edit] Using 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 $\mathbf{x}$ be an arbitrary real m-dimensional column vector such that || $\mathbf{x}$ || = |α| for a scalar α. If the algorithm is implemented using floating-point arithmetic, then α should get the opposite sign as the first coordinate of $\mathbf{x}$ to avoid loss of significance. In the complex case, set

$\alpha = - \mathrm{e}^{\mathrm{i} \arg x_1} \|\mathbf{x}\|$

(Stoer & Bulirsch 2002, p. 225) and substitute transposition by conjugate transposition in the construction of Q below.

Then, where $\mathbf{e}_1$ is the vector (1,0,...,0)^T, and ||·|| the Euclidean norm, set

$\mathbf{u} = \mathbf{x} - \alpha\mathbf{e}_1,$

$\mathbf{v} = {\mathbf{u}\over\|\mathbf{u}\|},$

$Q = I - 2 \mathbf{v}\mathbf{v}^T.$

$Q$ is a Householder matrix and

$Qx = (\alpha, 0, \cdots, 0)^T.\,$

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 Q₁A with zeros in the left column (except for the first row).

$Q_1A = \begin{bmatrix} \alpha_1&\star&\dots&\star\\ 0 & & & \\ \vdots & & A' & \\ 0 & & & \end{bmatrix}$

This can be repeated for A′ (obtained from Q₁A 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:

$Q_k = \begin{pmatrix} I_{k-1} & 0\\ 0 & Q_k'\end{pmatrix}.$

After $t$ iterations of this process, $t = min(m - 1, n)$ ,

$R = Q_t \cdots Q_2Q_1A$

is a upper triangular matrix. So, with

$Q = Q_1 Q_2 \cdots Q_t,$

$A = Q R$ is a QR decomposition of $A$ .

This method has greater numerical stability than the Gram-Schmidt method above.

The following table gives the number of operations in the k-th step of the QR-Decomposition by the Householder transformation, assuming a square matrix with size n.

Operation	Number of operations in the k-th step
multiplications	$2(n - k + 1) 2$
additions	$(n - k + 1) 2 + (n - k + 1)(n - k) + 2$
division	$1$
square root	$1$

Summing these numbers over the $(n - 1)$ steps (for a square matrix of size n), the complexity of the algorithm (in terms of floating point multiplications) is given by

$\frac{2}{3}n^3+n^2+\frac{1}{3}n-2=O(n^3)$

[edit] Example

Let us calculate the decomposition of

$A = \begin{pmatrix} 12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix}.$

First, we need to find a reflection that transforms the first column of matrix A, vector $\mathbf{a}_1 = (12, 6, -4)^T$ , to $\|\mathbf{a}_1\| \;\mathrm{e}_1 = (14, 0, 0)^T.$

Now,

$\mathbf{u} = \mathbf{x} - \alpha\mathbf{e}_1,$

and

$\mathbf{v} = {\mathbf{u}\over\|\mathbf{u}\|}.$

Here,

α = 14

and $\mathbf{x} = \mathbf{a}_1 = (12, 6, -4)^T$

Therefore

$\mathbf{u} = (-2, 6, -4)^T$ and $\mathbf{v} = {1 \over \sqrt{14}}(-1, 3, -2)^T$ , and then

$Q_1 = I - {2 \over \sqrt{14} \sqrt{14}} \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix}\begin{pmatrix} -1 & 3 & -2 \end{pmatrix}$

$= I - {1 \over 7}\begin{pmatrix} 1 & -3 & 2 \\ -3 & 9 & -6 \\ 2 & -6 & 4 \end{pmatrix}$

$= \begin{pmatrix} 6/7 & 3/7 & -2/7 \\ 3/7 &-2/7 & 6/7 \\ -2/7 & 6/7 & 3/7 \\ \end{pmatrix}.$

Now observe:

$Q_1A = \begin{pmatrix} 14 & 21 & -14 \\ 0 & -49 & -14 \\ 0 & 168 & -77 \end{pmatrix},$

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

$A' = M_{11} = \begin{pmatrix} -49 & -14 \\ 168 & -77 \end{pmatrix}.$

By the same method as above, we obtain the matrix of the Householder transformation

$Q_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -7/25 & 24/25 \\ 0 & 24/25 & 7/25 \end{pmatrix}$

after performing a direct sum with 1 to make sure the next step in the process works properly.

Now, we find

$Q=Q_1 Q_2=\begin{pmatrix} 6/7 & -69/175 & 58/175\\ 3/7 & 158/175 & -6/175 \\ -2/7 & 6/35 & 33/35 \end{pmatrix}$

$R=Q_2Q_1A=Q^T A=\begin{pmatrix} 14 & 21 & -14 \\ 0 & 175 & -70 \\ 0 & 0 & -35 \end{pmatrix}.$

The matrix Q is orthogonal and R is upper triangular, so A = QR is the required QR-decomposition.

[edit] Using 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.

[edit] Example

Let us calculate the decomposition of

$A = \begin{pmatrix} 12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix}.$

First, we need to form a rotation matrix that will zero the lowermost left element, $\mathbf{a}_{31} = -4$ . We form this matrix using the Givens rotation method, and call the matrix $G 1$ . We will first rotate the vector $(6, - 4)$ , to point along the X axis. This vector has an angle $\theta = \arctan({-4 \over 6})$ . We create the orthogonal Givens rotation matrix, $G 1$ :

$G_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) \\ 0 & -\sin(\theta) & \cos(\theta) \end{pmatrix}$

$\approx \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0.83205 & -0.55470 \\ 0 & 0.55470 & 0.83205 \end{pmatrix}$

And the result of $G 1 A$ now has a zero in the $\mathbf{a}_{31}$ element.

$G_1A \approx \begin{pmatrix} 12 & -51 & 4 \\ 7.21110 & 125.6396 & -33.83671 \\ 0 & 112.6041 & -71.83368 \end{pmatrix}$

We can similarly form Givens matrices $G 2$ and $G 3$ , which will zero the sub-diagonal elements $a 21$ and $a 32$ , forming a triangular matrix $R$ . The orthogonal matrix $Q T$ is formed from the concatenation of all the Givens matrices $Q T = G 3 G 2 G 1$ . Thus, we have $G 3 G 2 G 1 A = Q T A = R$ , and the QR decomposition is $A = Q R$ .

[edit] Connection to a determinant or a product of eigenvalues

We can use QR decomposition to find the absolute value of the determinant of a square matrix. Suppose a matrix is decomposed as $A = Q R$ . Then we have

$\det(A)=\det(Q)\cdot\det(R).$

Since Q is unitary, $| det(Q) | = 1$ . Thus,

$|\det(A)|=|\det(R)|=\Big|\prod_{i} r_{ii}\Big|,$

where $r i i$ are the entries on the diagonal of R.

Furthermore, because the determinant equals the product of the eigenvalues, we have

$\Big|\prod_{i} r_{ii}\Big|=\Big|\prod_{i} \lambda_{i}\Big|,$

where $λ i$ are eigenvalues of $A$ .

We can extend the above properties to non-square complex matrix $A$ by introducing the definition of QR-decomposition for non-square complex matrix and replacing eigenvalues with singular values.

Suppose a QR decomposition for a non-square matrix A:

$A = Q \begin{pmatrix}R\\O\end{pmatrix}, \qquad Q^*Q = I,$

where $O$ is a zero matrix and $Q$ is an unitary matrix.

From the properties of SVD and determinant of matrix, we have

$\Big|\prod_{i} r_{ii}\Big| = \prod_{i} \sigma_{i},$

where $σ i$ are singular values of $A$ .

Note that the singular values of $A$ and $R$ are identical, although the complex eigenvalues of them may be different. However, if A is square, it holds that

${\prod_{i} \sigma_{i}} = \Big|{\prod_{i} \lambda_{i}}\Big|.$

In conclusion, QR decomposition can be used efficiently to calculate a product of eigenvalues or singular values of matrix.

[edit] See also

[edit] References

Golub, Gene H.; Van Loan, Charles F. (1996), Matrix Computations (3rd ed.), Johns Hopkins, ISBN 978-0-8018-5414-9 .
Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN 0-521-38632-2 . Section 2.8.
Stoer, Josef; Bulirsch, Roland (2002), Introduction to Numerical Analysis (3rd ed.), Springer, ISBN 0-387-95452-X .

[edit] External links

Online Matrix Calculator Performs QR decomposition of matrices.
LAPACK users manual gives details of subroutines to calculate the QR decomposition
Mathematica users manual gives details and examples of routines to calculate QR decomposition
ALGLIB includes a partial port of the LAPACK to C++, C#, Delphi, etc.

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QR decomposition

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Contents

[edit] Definition

[edit] Computing the QR decomposition

[edit] Using the Gram-Schmidt process

[edit] Example

[edit] Relation to RQ decomposition

[edit] Application to computer vision

[edit] Using Householder reflections

[edit] Example

[edit] Using Givens rotations

[edit] Example

[edit] Connection to a determinant or a product of eigenvalues

[edit] See also

[edit] References

[edit] External links