91Ó°ÊÓ

Consider the matrix M as follows.

$\mathrm{M}=\frac{1}{2}\left[\begin{array}{ccc}1& 1& 1\\ 1& -1& -1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right]\left[\begin{array}{ccc}2& 3& 5\\ 0& -4& 6\\ 0& 0& 7\end{array}\right]$

Now, to find the QR-factorization.

Let’s write $||{\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{i}}||=1,i=1,2,3,{\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{i}}-{\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{j}}=0,\mathrm{i}≠\mathrm{j}$.

Here Q’ has orthonormal column and R’ is also an upper triangular matrix.

According to the theorem 5.2.2 any matrix with linearly independent column localid="1660102237605" ${\stackrel{â‡\P Ä}{\mathrm{v}}}_1,...,{\stackrel{â‡\P Ä}{\mathrm{v}}}_{\mathrm{m}}$can be expressed as follows.

localid="1660102271132" $\mathrm{M}=\mathrm{QR}\mathrm{where}\mathrm{Q}=\left[{\stackrel{â‡\P Ä}{\mathrm{u}}}_1{\stackrel{â‡\P Ä}{\mathrm{u}}}_2...{\stackrel{â‡\P Ä}{\mathrm{u}}}_{\mathrm{m}}\right],\mathrm{R}=$upper triangular matrix with positive diagonal entries and ${\stackrel{â‡\P Ä}{\mathrm{u}}}_1{\stackrel{â‡\P Ä}{\mathrm{u}}}_2...{\stackrel{â‡\P Ä}{\mathrm{u}}}_{\mathrm{m}}$are orthonormal.

${\mathrm{r}}_{11}=||{\stackrel{â‡\P Ä}{\mathrm{v}}}_1||,{\mathrm{r}}_{\mathrm{jj}}=||{\mathrm{v}}_{\mathrm{j}}^{âŠ\yen }||=||{\mathrm{v}}_{\mathrm{j}}-\underset{\mathrm{k}=1}{\overset{\mathrm{j}-1}{∑}}\left({\stackrel{â‡\P Ä}{\mathrm{u}}}_{\mathrm{k}}-{\stackrel{â‡\P Ä}{\mathrm{v}}}_{\mathrm{j}}\right){\stackrel{â‡\P Ä}{\mathrm{u}}}_{\mathrm{k}}||,{\mathrm{r}}_{\mathrm{ij}}={\stackrel{â‡\P Ä}{\mathrm{v}}}_{\mathrm{i}}-{\stackrel{â‡\P Ä}{\mathrm{v}}}_{\mathrm{j}}$

As in the QR factorization the diagonal entries of R should be positive.

So, the only entry in R’ which is creating problem is ${\mathrm{r}}_{22}=-4$.

In order to change the sign in the diagonal without changing the matrix and only need to change the sign in the second column for Q’ and the second for R’.

Thus, the QR factorization is

$$\begin{gathered} \mathrm{Q}=\frac{1}{2}\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ 1& 1& 1\\ 1& -1& -1\end{array}\right] \\ \mathrm{R}=\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& -6\\ 0& 0& 7\end{array}\right] \\ \mathrm{M}=\frac{1}{2}\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ 1& 1& 1\\ 1& -1& -1\end{array}\right]\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& -6\\ 0& 0& 7\end{array}\right] \end{gathered}$$