Editing QR decomposition (section)

====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>.}}