91Ó°ÊÓ

In QR factorization of a matrix A, the matrix is decomposed into a product A = QR, where Q is an orthonormal matrix and R is an upper triangular matrix.

The given matrix is $M=\frac{1}{2}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& -1& 1\\ 1& -1& 1& -1\\ 1& 1& -1& -1\end{array}\right]\left[\begin{array}{cc}3& 4\\ 0& 5\\ 0& 0\\ 0& 0\end{array}\right]$. Simplify the given matrix to get $M=\frac{1}{2}\left[\begin{array}{cc}3& 9\\ 3& -1\\ 3& -1\\ 3& 9\end{array}\right]$.

Now, find a matrix Q with the help of orthonormal basis for the ColM. Suppose the columns of the matrix M are $v_1=\frac{1}{2}\left[\begin{array}{c}3\\ 3\\ 3\\ 3\end{array}\right]$and $v_2=\frac{1}{2}\left[\begin{array}{c}9\\ -1\\ -1\\ 9\end{array}\right]$.

Now, the orthonormal basis for$\left\{v_1,v_2\right\}$ can be obtained using the Gram Schmidt process:

$\begin{array}{rcl}{u}_1& =& \frac{1}{\left|\left|\stackrel{â‡\P Ä}{v}\right|\right|}\stackrel{â‡\P Ä}{v_1}\\ & =& \frac{1}{2}\left(\frac{1}{2}\left[\begin{array}{c}3\\ 3\\ 3\\ 3\end{array}\right]\right)\\ & =& \frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\end{array}$

Compute u₂ usingrole="math" localid="1660299207883" $u_2=\frac{{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }}{\left|\left|{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }\right|\right|}$ and role="math" localid="1660298868884" ${{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }={\stackrel{â‡\P Ä}{v}}_2-\left({\stackrel{â‡\P Ä}{u}}_1·{\stackrel{â‡\P Ä}{v}}_2\right){\stackrel{â‡\P Ä}{u}}_1$:

$\begin{array}{rcl}{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }& =& {\stackrel{â‡\P Ä}{v}}_2-\left({\stackrel{â‡\P Ä}{u}}_1·{\stackrel{â‡\P Ä}{v}}_2\right){\stackrel{â‡\P Ä}{u}}_1\\ & =& \frac{1}{2}\left[\begin{array}{c}9\\ -1\\ -1\\ 9\end{array}\right]-\frac{4}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\\ & =& \frac{1}{2}\left[\begin{array}{c}9\\ -1\\ -1\\ 9\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right]\\ & =& \frac{1}{2}\left[\begin{array}{c}5\\ -5\\ -5\\ 5\end{array}\right]\end{array}$

So, the vector u₂ is:

$\begin{array}{rcl}{u}_2& =& \frac{{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }}{\left|\left|{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }\right|\right|}\\ & =& \frac{1}{5}\left(\frac{1}{2}\left[\begin{array}{c}5\\ -5\\ -5\\ 5\end{array}\right]\right)\\ & =& \frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\end{array}$

Now, the orthonormal matrix Q is [u₁ u₂] which is given below:

$Q=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& -1\\ 1& -1\\ 1& 1\end{array}\right]$

Now, the upper triangular matrix R can be obtained by using the formularole="math" localid="1660299820552" $R=\left[\begin{array}{cc}\left|\left|v_1\right|\right|& {\stackrel{â‡\P Ä}{u}}_1·{\stackrel{â‡\P Ä}{v}}_2\\ 0& \left|\left|{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }\right|\right|\end{array}\right]$ as follows:

role="math" localid="1660299880053" $\begin{array}{rcl}R& =& \left[\begin{array}{cc}\left|\left|v_1\right|\right|& {\stackrel{â‡\P Ä}{u}}_1·{\stackrel{â‡\P Ä}{v}}_2\\ 0& \left|\left|{{\stackrel{â‡\P Ä}{v}}_2}^{âŠ\yen }\right|\right|\end{array}\right]\\ & =& \left[\begin{array}{cc}3& 4\\ 0& 5\end{array}\right]\end{array}$

So, the upper triangular matrix is $R\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \left[\begin{array}{cc}3& 4\\ 0& 5\end{array}\right]\end{array}$. The QR factorization of the matrix M is given by role="math" localid="1660300352960" $QR=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& -1\\ 1& -1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 4\\ 0& 5\end{array}\right]$.

Thus, the QR factorization of the given matrix M is $\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& -1\\ 1& -1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 4\\ 0& 5\end{array}\right]$.