# All Science Fair Projects

## Science Fair Project Encyclopedia for Schools!

 Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary

# Science Fair Project Encyclopedia

For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.

# QR decomposition

In linear algebra, the QR decomposition of a matrix A is a factorization expressing A as

A = QR

where Q is an orthogonal matrix (QQT = I), and R 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.)

 Contents

## 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 $A=(\mathbf{a}_1| \cdots|\mathbf{a}_n)$. Then

$\mathbf{u}_1 = \mathbf{a}_1, \qquad\mathbf{e}_1 = {\mathbf{u}_1 \over ||\mathbf{u}_1||}$
$\mathbf{u}_2 = \mathbf{a}_2-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2, \qquad\mathbf{e}_2 = {\mathbf{u}_2 \over ||\mathbf{u}_2||}$
$\mathbf{u}_3 = \mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3-\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3, \qquad\mathbf{e}_3 = {\mathbf{u}_3 \over ||\mathbf{u}_3||}$
$\vdots$
$\mathbf{u}_k = \mathbf{a}_k-\sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k,\qquad\mathbf{e}_k = {\mathbf{u}_k\over||\mathbf{u}_k||}$

Naturally then, we rearrange the equations so the ais are the subject, to get the following

$\mathbf{a}_1 = \mathbf{e}_1||\mathbf{u}_1||$
$\mathbf{a}_2 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_2+\mathbf{e}_2||\mathbf{u}_2||$
$\mathbf{a}_3 = \mathrm{proj}_{\mathbf{e}_1}\,\mathbf{a}_3+\mathrm{proj}_{\mathbf{e}_2}\,\mathbf{a}_3+\mathbf{e}_3||\mathbf{u}_3||$
$\vdots$
$\mathbf{a}_k = \sum_{j=1}^{k-1}\mathrm{proj}_{\mathbf{e}_j}\,\mathbf{a}_k+\mathbf{e}_k||\mathbf{u}_k||$

Each of these projections of the vectors $\mathbf{a}_i$ onto one of these ej are merely the inner product of the two, since the vectors are normed.

Now these equations can be written in matrix form, viz.,

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

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 eks), 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 QQT = I, so QT = Q - 1.

### Example

Consider the decomposition of

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

Recall the orthogonal matrix Q such that

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

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

$U = \begin{pmatrix} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{pmatrix} = \begin{pmatrix} 12 & -51 & 4 \\ 6 & 167 & -68 \\ -4 & 24 & -41 \end{pmatrix};$
$Q = \begin{pmatrix} \frac{\mathbf u_1}{||\mathbf u_1||} & \frac{\mathbf u_2}{||\mathbf u_2||} & \frac{\mathbf u_3}{||\mathbf u_3||} \end{pmatrix} = \begin{pmatrix} 6/7 & -69/175 & -58/175 \\ 3/7 & 158/175 & 6/175 \\ -2/7 & 6/35 & -33/35 \end{pmatrix};$

Thus, we have

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

Considering numerical errors of finite precision operation in MATLAB, we have that

$\begin{matrix} Q = \end{matrix} \begin{pmatrix} 0.857142857142857 & -0.394285714285714 & -0.331428571428571 \\ 0.428571428571429 & 0.902857142857143 & 0.034285714285714 \\ -0.285714285714286 & 0.171428571428571 & -0.942857142857143 \end{pmatrix};$
$\begin{matrix} R = \end{matrix} \begin{pmatrix} 14 & 21 & -14 \\ 1.11022302462516 \times 10^{-016} & 175 & -70 \\ -1.77635683940025 \times 10^{-015} & -5.32907051820075 \times 10^{-014} & 35 \end{pmatrix}.$

## 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

$\mathbf{u} = \mathbf{x} - \alpha\mathbf{e}_1,$
$\mathbf{v} = {\mathbf{u}\over||\mathbf{u}||},$
$Q = I - 2 \mathbf{v}\mathbf{v}^T.$.

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 QA with zeros in the left column (except for the first line).

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

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:

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

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

$Q = Q_1Q_2 \cdots Q_t$

A = QR is a QR decomposition of A.

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

### Example

Let us calculate the decomposition of

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

We need to find a reflection that takes the vector a1 = (12,6, - 4)T to $\| \;\mathrm{a}\| \;\mathrm{e}_1 = (14, 0, 0)^T$.

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

$Q_1 = I - {2 \over 14} \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix}\begin{pmatrix} -1 & 3 & -2 \end{pmatrix}$
$= I - {1 \over 7}\begin{pmatrix} 1 & -3 & 2 \\ -3 & 9 & -6 \\ 2 & -6 & 4 \end{pmatrix} =\begin{pmatrix} 6/7 & 3/7 & -2/7 \\ 3/7 &-2/7 & 6/7 \\ -2/7 & 6/7 & 3/7 \\ \end{pmatrix}$

Observe now:

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

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 to be,

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

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

Now, we find

$Q=Q_1Q_2=\begin{pmatrix} 6/7 & -69/175 & 58/175 \\ 3/7 & 158/175 & -6/175 \\ -2/7 & 6/35 & 33/35 \end{pmatrix}$
$R=Q^\top A=\begin{pmatrix} 14 & 21 & -14 \\ 0 & 175 & -70 \\ 0 & 0 & -35 \end{pmatrix}$

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

03-10-2013 05:06:04