Editing QR decomposition (section)

==Using for solution to linear inverse problems==
Compared to the direct matrix inverse, inverse solutions using QR decomposition are more numerically stable as evidenced by their reduced [[condition number]]s.<ref>{{Cite book |last=Parker |first=Robert L. |url=https://www.worldcat.org/oclc/1134769155 |title=Geophysical Inverse Theory |date=1994 |publisher=Princeton University Press |isbn=978-0-691-20683-7 | location=Princeton, N.J. |oclc=1134769155 | at = Section 1.13 }}</ref>

To solve the underdetermined {{nowrap|(<math>m < n</math>)}} linear problem <math>A \mathbf x = \mathbf b</math> where the matrix <math>A</math> has dimensions <math>m \times n</math> and rank {{nowrap|<math>m</math>,}} first find the QR factorization of the transpose of {{nowrap|<math>A</math>:}} {{nowrap|<math>A^\textsf{T} = QR</math>,}} where ''Q'' is an orthogonal matrix (i.e. {{nowrap|<math>Q^\textsf{T} = Q^{-1}</math>),}} and ''R'' has a special form: <math>R = \left[\begin{smallmatrix} R_1 \\ 0 \end{smallmatrix}\right]</math>. Here <math>R_1</math> is a square <math>m \times m</math> right triangular matrix, and the zero matrix has dimension {{nowrap|<math>(n-m) \times m</math>.}} After some algebra, it can be shown that a solution to the inverse problem can be expressed as: <math>\mathbf x = Q \left[\begin{smallmatrix}
\left(R_1^\textsf{T}\right)^{-1} \mathbf b \\
                0
\end{smallmatrix}\right]</math> where one may either find <math>R_1^{-1}</math> by [[Gaussian elimination]] or compute <math>\left(R_1^\textsf{T}\right)^{-1} \mathbf b</math> directly by [[triangular matrix#Forward and back substitution|forward substitution]]. The latter technique enjoys greater numerical accuracy and lower computations.

To find a solution <math>\hat{\mathbf x}</math> to the overdetermined {{nowrap|(<math>m \geq n</math>)}} problem <math>A \mathbf x = \mathbf b</math> which minimizes the norm {{nowrap|<math>\left\|A \hat{\mathbf{x}} - \mathbf{b}\right\|</math>,}} first find the QR factorization of {{nowrap|<math>A</math>:}} {{nowrap|<math>A = QR</math>.}} The solution can then be expressed as {{nowrap|<math>\hat{\mathbf x} = R_1^{-1} \left(Q_1^\textsf{T} \mathbf{b}\right) </math>,}} where <math>Q_1</math> is an <math>m \times n</math> matrix containing the first <math>n</math> columns of the full orthonormal basis <math>Q</math> and where <math>R_1</math> is as before. Equivalent to the underdetermined case, [[triangular matrix#Forward and back substitution|back substitution]] can be used to quickly and accurately find this <math>\hat{\mathbf{x}}</math> without explicitly inverting {{nowrap|<math>R_1</math>.}} (<math>Q_1</math> and <math>R_1</math> are often provided by numerical libraries as an "economic" QR decomposition.)