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 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.