Editing QR decomposition (section)

==Computing the QR decomposition==
There are several methods for actually computing the QR decomposition, such as the [[Gram–Schmidt process]], [[Householder transformation]]s, or [[Givens rotation]]s. Each has a number of advantages and disadvantages.

===Using the Gram–Schmidt process===
{{further|Gram–Schmidt#Numerical stability}}
Consider the [[Gram–Schmidt process]] applied to the columns of the full column rank matrix {{nowrap|<math>A = \begin{bmatrix}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{bmatrix}</math>,}} with [[inner product]] <math>\langle\mathbf{v}, \mathbf{w}\rangle = \mathbf{v}^\textsf{T} \mathbf{w}</math> (or <math>\langle\mathbf{v}, \mathbf{w}\rangle = \mathbf{v}^\dagger \mathbf{w}</math> for the complex case).

Define the [[vector projection|projection]]:
:<math>\operatorname{proj}_{\mathbf{u}}\mathbf{a} =
  \frac{\left\langle\mathbf{u}, \mathbf{a}\right\rangle}{\left\langle\mathbf{u}, \mathbf{u}\right\rangle}{\mathbf{u}}
</math>

then:
:<math>\begin{align}
  \mathbf{u}_1 &= \mathbf{a}_1, &
    \mathbf{e}_1 &= \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} \\
  \mathbf{u}_2 &= \mathbf{a}_2 - \operatorname{proj}_{\mathbf{u}_1} \mathbf{a}_2, &
    \mathbf{e}_2 &= \frac{\mathbf{u}_2}{\|\mathbf{u}_2\|} \\
  \mathbf{u}_3 &= \mathbf{a}_3 - \operatorname{proj}_{\mathbf{u}_1} \mathbf{a}_3 - \operatorname{proj}_{\mathbf{u}_2} \mathbf{a}_3, &
    \mathbf{e}_3 &= \frac{\mathbf{u}_3}{\|\mathbf{u}_3\|} \\
                 & \;\; \vdots &  & \;\; \vdots \\
  \mathbf{u}_k &= \mathbf{a}_k - \sum_{j=1}^{k-1}\operatorname{proj}_{\mathbf{u}_j} \mathbf{a}_k,&
    \mathbf{e}_k &= \frac{\mathbf{u}_k}{\|\mathbf{u}_k\|}
\end{align}</math>

We can now express the <math>\mathbf{a}_i</math>s over our newly computed orthonormal basis:
:<math>\begin{align}
   \mathbf{a}_1 &= \left\langle\mathbf{e}_1, \mathbf{a}_1\right\rangle \mathbf{e}_1 \\
   \mathbf{a}_2 &= \left\langle\mathbf{e}_1, \mathbf{a}_2\right\rangle \mathbf{e}_1
                 + \left\langle\mathbf{e}_2, \mathbf{a}_2\right\rangle \mathbf{e}_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
                 + \left\langle\mathbf{e}_3, \mathbf{a}_3\right\rangle \mathbf{e}_3 \\
                &\;\;\vdots \\
   \mathbf{a}_k &= \sum_{j=1}^k \left\langle \mathbf{e}_j, \mathbf{a}_k \right\rangle \mathbf{e}_j
\end{align}</math>

where {{nowrap|<math>\left\langle\mathbf{e}_i, \mathbf{a}_i\right\rangle = \left\|\mathbf{u}_i\right\|</math>.}} This can be written in matrix form:
:<math>A = QR</math>

where:
:<math>Q = \begin{bmatrix}\mathbf{e}_1 & \cdots & \mathbf{e}_n\end{bmatrix}</math>

and
:<math>R = \begin{bmatrix}
  \langle\mathbf{e}_1, \mathbf{a}_1\rangle &
  \langle\mathbf{e}_1, \mathbf{a}_2\rangle &
  \langle\mathbf{e}_1, \mathbf{a}_3\rangle & 
                                    \cdots &
  \langle\mathbf{e}_1, \mathbf{a}_n\rangle \\
                                         0 &
  \langle\mathbf{e}_2, \mathbf{a}_2\rangle &
  \langle\mathbf{e}_2, \mathbf{a}_3\rangle & 
                                   \cdots &
  \langle\mathbf{e}_2, \mathbf{a}_n\rangle \\
                                         0 &
                                         0 &
  \langle\mathbf{e}_3, \mathbf{a}_3\rangle & 
                                    \cdots & 
  \langle\mathbf{e}_3, \mathbf{a}_n\rangle \\
                                    \vdots &
                                    \vdots &
                                    \vdots &
                                    \ddots &
                                    \vdots \\
                                         0 &
                                         0 &
                                         0 &
                                    \cdots &
 \langle\mathbf{e}_n, \mathbf{a}_n\rangle \\
                                    
\end{bmatrix}.</math>

====Example====
Consider the decomposition of
: <math>A = \begin{bmatrix}
  12 & -51 &   4 \\
   6 & 167 & -68 \\
  -4 &  24 & -41
\end{bmatrix}.</math>

Recall that an orthonormal matrix <math>Q</math> has the property {{nowrap|<math>Q^\textsf{T} Q = I</math>.}}

Then, we can calculate <math>Q</math> by means of Gram–Schmidt as follows:
: <math>\begin{align}
  U = \begin{bmatrix} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{bmatrix}
   &= \begin{bmatrix}
        12 & -69 & -58/5 \\
         6 & 158 &   6/5 \\
        -4 &  30 & -33
      \end{bmatrix}; \\
  Q = \begin{bmatrix}
        \frac{\mathbf u_1}{\|\mathbf u_1\|} &
        \frac{\mathbf u_2}{\|\mathbf u_2\|} &
        \frac{\mathbf u_3}{\|\mathbf u_3\|}
      \end{bmatrix} &=
      \begin{bmatrix}
         6/7 & -69/175 & -58/175 \\
         3/7 & 158/175 &   6/175 \\
        -2/7 &   6/35  & -33/35
      \end{bmatrix}.
\end{align}</math>

Thus, we have
: <math>\begin{align}
  Q^\textsf{T} A &= Q^\textsf{T}Q\,R = R; \\
               R &= Q^\textsf{T}A =
    \begin{bmatrix}
      14 &  21 & -14 \\
       0 & 175 & -70 \\
       0 &   0 &  35
    \end{bmatrix}.
\end{align}</math>

====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.

====Advantages and disadvantages====

The Gram-Schmidt process is inherently numerically unstable. While the application of the projections has an appealing geometric analogy to orthogonalization, the orthogonalization itself is prone to numerical error. A significant advantage is the ease of implementation.

===Using Householder reflections===

{{see also|Householder transformation}}

[[File:Householder.svg|thumb|Householder reflection for QR-decomposition: The goal is to find a linear transformation that changes the vector <math>\mathbf x</math> into a vector of the same length which is collinear to <math>\mathbf e_1</math>. We could use an orthogonal projection (Gram-Schmidt) but this will be numerically unstable if the vectors <math>\mathbf x</math> and <math>\mathbf e_1</math> are close to orthogonal. Instead, the Householder reflection reflects through the dotted line (chosen to bisect the angle between <math>\mathbf x</math> and {{nowrap|<math>\mathbf e_1</math>).}} The maximum angle with this transform is 45 degrees.]]

A Householder reflection (or ''Householder transformation'') is a transformation that takes a vector and reflects it about some [[plane (mathematics)|plane]] or [[hyperplane]]. We can use this operation to calculate the ''QR'' factorization of an ''m''-by-''n'' matrix <math>A</math> with {{nowrap|''m'' ≥ ''n''}}.

''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 real ''m''-dimensional column vector of <math>A</math> such that <math>\|\mathbf{x}\| = |\alpha|</math> for a scalar ''α''. If the algorithm is implemented using [[floating-point arithmetic]], then ''α'' should get the opposite sign as the ''k''-th coordinate of {{nowrap|<math>\mathbf{x}</math>,}} where <math>x_k</math> is to be the pivot coordinate after which all entries are 0 in matrix ''A''{{'}}s final upper triangular form, to avoid [[loss of significance]]. In the complex case, set<ref>{{citation | first1=Josef | last1=Stoer | first2=Roland | last2=Bulirsch | year=2002 | title=Introduction to Numerical Analysis | edition=3rd | publisher=Springer | isbn=0-387-95452-X |page=225}}</ref>
:<math>\alpha = -e^{i \arg x_k} \|\mathbf{x}\|</math>
and substitute transposition by conjugate transposition in the construction of ''Q'' below.

Then, where <math>\mathbf{e}_1</math> is the vector {{math|[1 0 ⋯ 0]<sup>T</sup>}}, {{math|{{!}}{{!}} · {{!}}{{!}}}} is the [[Euclidean space#Euclidean norm|Euclidean norm]] and <math>I</math> is an {{math|''m''×''m''}} identity matrix, set
: <math>\begin{align}
  \mathbf{u} &= \mathbf{x} - \alpha\mathbf{e}_1, \\
  \mathbf{v} &= \frac{\mathbf{u}}{\|\mathbf{u}\|}, \\
           Q &= I - 2 \mathbf{v}\mathbf{v}^\textsf{T}.
\end{align}</math>

Or, if <math>A</math> is complex
: <math>Q = I - 2\mathbf{v}\mathbf{v}^\dagger.</math>

<math>Q</math> is an ''m''-by-''m'' Householder matrix, which is both symmetric and orthogonal (Hermitian and unitary in the complex case), and
: <math>Q\mathbf{x} = \begin{bmatrix} \alpha \\ 0 \\ \vdots \\ 0 \end{bmatrix}.</math>

This can be used to gradually transform an ''m''-by-''n'' matrix ''A'' to upper [[Triangular matrix|triangular]] form. First, we multiply ''A'' with the Householder matrix ''Q''<sub>1</sub> we obtain when we choose the first matrix column for '''x'''. This results in a matrix ''Q''<sub>1</sub>''A'' with zeros in the left column (except for the first row).
: <math>Q_1A = \begin{bmatrix}
  \alpha_1 & \star & \cdots & \star \\
         0 &       &        &       \\
    \vdots &       &     A' &       \\
         0 &       &        &
\end{bmatrix}</math>

This can be repeated for ''A''′ (obtained from ''Q''<sub>1</sub>''A'' by deleting the first row and first column), resulting in a Householder matrix ''Q''′<sub>2</sub>. Note that ''Q''′<sub>2</sub> is smaller than ''Q''<sub>1</sub>. Since we want it really to operate on ''Q''<sub>1</sub>''A'' instead of ''A''′ we need to expand it to the upper left, filling in a 1, or in general:
:<math>Q_k = \begin{bmatrix}
  I_{k-1} & 0    \\
       0  & Q_k'
\end{bmatrix}.</math>

After <math>t</math> iterations of this process, {{nowrap|<math>t = \min(m - 1, n)</math>,}}
:<math>R = Q_t \cdots Q_2 Q_1 A</math>

is an upper triangular matrix. So, with
:<math>\begin{align}
Q^\textsf{T} &= Q_t \cdots Q_2 Q_1, \\
Q &= Q_1^\textsf{T} Q_2^\textsf{T} \cdots Q_t^\textsf{T}
\end{align}</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.<!--See the below example, and compare above-->

In numerical tests the computed factors <math>Q_c</math> and <math>R_c</math> satisfy
<math>\frac{\|Q R - Q_c R_c\|_\infty}{\|A\|_\infty} = O(\varepsilon)</math>
at machine precision. Also, orthogonality is preserved: <math>\|Q_c^\mathsf{T} Q_c - I\|_\infty = O(\varepsilon)</math>. However, the accuracy of <math>Q_c</math> and <math>R_c</math> decrease with condition number:
<math>\|Q - Q_c\|_\infty = O(\varepsilon\,\kappa_\infty(A)),\quad
\frac{\|R - R_c\|_\infty}{\|R\|_\infty} = O(\varepsilon\,\kappa_\infty(A)).</math>

For a well-conditioned example (<math>n=4000</math>, <math>\kappa_\infty(A)\approx3\times10^{3}</math>):
<math>\frac{\|Q R - Q_c R_c\|_\infty}{\|A\|_\infty} \approx 1.6\times10^{-15},</math>
<math>\|Q - Q_c\|_\infty \approx 1.6\times10^{-15},</math>
<math>\frac{\|R - R_c\|_\infty}{\|R\|_\infty} \approx 4.3\times10^{-14},</math>
<math>\|Q_c^\mathsf{T}Q_c - I\|_\infty \approx 1.1\times10^{-13}.</math>

In an ill-conditioned test (<math>n=4000</math>, <math>\kappa_\infty(A)\approx4\times10^{18}</math>):
<math>\frac{\|Q R - Q_c R_c\|_\infty}{\|A\|_\infty} \approx 1.3\times10^{-15},</math>
<math>\|Q - Q_c\|_\infty \approx 5.2\times10^{-4},</math>
<math>\frac{\|R - R_c\|_\infty}{\|R\|_\infty} \approx 1.2\times10^{-4},</math>
<math>\|Q_c^\mathsf{T}Q_c - I\|_\infty \approx 1.1\times10^{-13}.</math><ref>{{Cite book | author1=Holmes, M.| title=Introduction to Scientific Computing and Data Analysis, 2nd Ed | year=2023 | publisher=Springer | isbn=978-3-031-22429-4}} </ref>

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''.
{| class="wikitable"
|-
! Operation
! Number of operations in the ''k''-th step
|-
| Multiplications
| <math>2(n - k + 1)^2</math>
|-
| Additions
| <math>(n - k + 1)^2 + (n - k + 1)(n - k) + 2 </math>
|-
| Division
| <math>1</math>
|-
| Square root
| <math>1</math>
|}

Summing these numbers over the {{nowrap|''n'' − 1}} steps (for a square matrix of size ''n''), the complexity of the algorithm (in terms of floating point multiplications) is given by
:<math>\frac{2}{3}n^3 + n^2 + \frac{1}{3}n - 2 = O\left(n^3\right).</math>

====Example====
Let us calculate the decomposition of
: <math>A = \begin{bmatrix}
  12 & -51 &   4 \\
   6 & 167 & -68 \\
  -4 &  24 & -41
\end{bmatrix}.</math>

First, we need to find a reflection that transforms the first column of matrix ''A'', vector {{nowrap|<math>\mathbf{a}_1 = \begin{bmatrix} 12 & 6 & -4 \end{bmatrix}^\textsf{T}</math>,}} into {{nowrap|<math>\left\|\mathbf{a}_1\right\| \mathbf{e}_1 = \begin{bmatrix} \alpha & 0 & 0\end{bmatrix}^\textsf{T}</math>.}}

Now,
: <math>\mathbf{u} = \mathbf{x} - \alpha\mathbf{e}_1,</math>

and
: <math>\mathbf{v} = \frac{\mathbf{u}}{\|\mathbf{u}\|}.</math>

Here,
: <math>\alpha = 14</math> and <math>\mathbf{x} = \mathbf{a}_1 = \begin{bmatrix} 12 & 6 & -4 \end{bmatrix}^\textsf{T}</math>

Therefore
: <math>\mathbf{u} = \begin{bmatrix} -2 & 6 & -4 \end{bmatrix}^\textsf{T} = 2 \begin{bmatrix} -1 & 3 & -2 \end{bmatrix}^\textsf{T}</math> and {{nowrap|<math>\mathbf{v} = \frac{1}{\sqrt{14}}\begin{bmatrix} -1 & 3 & -2 \end{bmatrix}^\textsf{T}</math>,}} and then
: <math>\begin{align}
      Q_1
  ={} &I - \frac{2}{\sqrt{14}\sqrt{14}}
         \begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix}
         \begin{bmatrix} -1 &  3 &  -2 \end{bmatrix} \\
  ={} &I - \frac{1}{7}\begin{bmatrix}
          1 & -3 &  2 \\
         -3 &  9 & -6 \\
          2 & -6 &  4
       \end{bmatrix} \\
  ={} &\begin{bmatrix}
          6/7 &  3/7 & -2/7 \\
          3/7 & -2/7 &  6/7 \\
         -2/7 &  6/7 &  3/7 \\
       \end{bmatrix}.
\end{align}</math>

Now observe:
:<math>Q_1A = \begin{bmatrix}
  14 &  21 & -14 \\
   0 & -49 & -14 \\
   0 & 168 & -77
\end{bmatrix},</math>

so we already have almost a triangular matrix. We only need to zero the (3, 2) entry.

Take the (1, 1) [[minor (linear algebra)|minor]], and then apply the process again to
:<math>A' = M_{11} = \begin{bmatrix}
  -49 & -14 \\
  168 & -77
\end{bmatrix}.</math>

By the same method as above, we obtain the matrix of the Householder transformation
:<math>Q_2 = \begin{bmatrix}
  1 &     0 &  0 \\
  0 & -7/25 & 24/25 \\
  0 & 24/25 &  7/25
\end{bmatrix}</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_1^\textsf{T} Q_2^\textsf{T} = \begin{bmatrix}
   6/7 & -69/175 & 58/175 \\
   3/7 & 158/175 & -6/175 \\
  -2/7 &   6/35  & 33/35
\end{bmatrix}. </math>

Or, to four decimal digits,
:<math>\begin{align}
  Q &= Q_1^\textsf{T} Q_2^\textsf{T} = \begin{bmatrix}
     0.8571 & -0.3943 &  0.3314 \\
     0.4286 &  0.9029 & -0.0343 \\
    -0.2857 &  0.1714 &  0.9429
  \end{bmatrix} \\
  R &= Q_2 Q_1 A = Q^\textsf{T} A = \begin{bmatrix}
    14 &  21 & -14 \\
     0 & 175 & -70 \\
     0 &   0 & -35
  \end{bmatrix}.
\end{align}</math>

The matrix ''Q'' is orthogonal and ''R'' is upper triangular, so {{nowrap|1=''A'' = ''QR''}} is the required QR decomposition.

====Advantages and disadvantages====

The use of Householder transformations is inherently the most simple of the numerically stable QR decomposition algorithms due to the use of reflections as the mechanism for producing zeroes in the ''R'' matrix. However, the Householder reflection algorithm is bandwidth heavy and difficult to parallelize, as every reflection that produces a new zero element changes the entirety of both ''Q'' and ''R'' matrices.

====Parallel implementation of Householder QR====

The Householder QR method can be implemented in parallel with algorithms such as the TSQR algorithm (which stands for ''Tall Skinny QR''). This algorithm can be applied in the case when the matrix ''A'' has ''m >> n''.<ref>Communication-optimal parallel and sequential QR and LU factorizations: theory and practice, James Demmel and Laura Grigori, 2008,
https://arxiv.org/abs/0806.2159, 
</ref> This algorithm uses a binary reduction tree to compute local householder QR decomposition at each node in the forward pass, and re-constitute the Q matrix in the backward pass. The [[binary tree]] structure aims at decreasing the amount of communication between processor to increase performance.

===Using Givens rotations===
QR decompositions can also be computed with a series of [[Givens rotation]]s. Each rotation zeroes 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 relatively few off-diagonal elements need to be zeroed, and is more easily parallelized than [[Householder transformation]]s.

====Example====
Let us calculate the decomposition of
: <math>A = \begin{bmatrix}
  12 & -51 &   4 \\
   6 & 167 & -68 \\
  -4 &  24 & -41
\end{bmatrix}.</math>

First, we need to form a [[rotation matrix]] that will zero the lowermost left element, {{nowrap|1=<math>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 {{nowrap|<math>\begin{bmatrix} 12 & -4 \end{bmatrix}</math>,}} to point along the ''X'' axis.  This vector has an angle {{nowrap|<math display="inline">\theta = \arctan\left(\frac{-(-4)}{12}\right)</math>.}}  We create the orthogonal Givens rotation matrix, <math>G_1</math>:

:<math>\begin{align}
  G_1 &= \begin{bmatrix}
    \cos(\theta) & 0 & -\sin(\theta) \\
               0 & 1 &             0 \\
    \sin(\theta) & 0 &  \cos(\theta)
  \end{bmatrix} \\
      &\approx \begin{bmatrix}
    0.94868 & 0 & -0.31622 \\
    0       & 1 &  0       \\
    0.31622 & 0 &  0.94868 
  \end{bmatrix}
\end{align}</math>

And the result of <math>G_1A</math> now has a zero in the <math>a_{31}</math> element.
:<math>G_1A \approx \begin{bmatrix}
  12.64911 & -55.97231 &  16.76007 \\
   6       & 167       & -68       \\
   0       &   6.64078 & -37.6311 
\end{bmatrix}</math>

We can similarly form Givens matrices <math>G_2</math> and {{nowrap|<math>G_3</math>,}} which will zero the sub-diagonal elements <math>a_{21}</math> and {{nowrap|<math>a_{32}</math>,}} forming a triangular matrix {{nowrap|<math>R</math>.}}  The orthogonal matrix <math>Q^\textsf{T}</math> is formed from the product of all the Givens matrices {{nowrap|<math>Q^\textsf{T} = G_3 G_2 G_1</math>.}}  Thus, we have {{nowrap|<math>G_3 G_2 G_1 A = Q^\textsf{T} A = R</math>,}} and the ''QR'' decomposition is {{nowrap|<math>A = QR</math>.}}

====Advantages and disadvantages====

The QR decomposition via Givens rotations is the most involved to implement, as the ordering of the rows required to fully exploit the algorithm is not trivial to determine. However, it has a significant advantage in that each new zero element <math>a_{ij}</math> affects only the row with the element to be zeroed (''i'') and a row above (''j''). This makes the Givens rotation algorithm more bandwidth efficient and parallelizable than the Householder reflection technique.