Question

Consider the matrix

$\mathit{M}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left[}\begin{array}{cccc}\mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{1}& \mathbf{-}\mathbf{1}\\ \mathbf{1}& \mathbf{1}& \mathbf{-}\mathbf{1}& \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{cc}\mathbf{3}& \mathbf{4}\\ \mathbf{0}& \mathbf{5}\\ \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}\end{array}\mathbf{\right]}$

Find the QR factorization of M.

Short answer

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

Step-by-step solution

factorization of matrix

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.

Find factorization of the given 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:

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

Compute u₂ usingrole="math" localid="1660299207883" $u_2=\frac{{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }}{\left|\left|{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }\right|\right|}$ and role="math" localid="1660298868884" ${{\stackrel{\rightharpoonup }{v}}_2}^{\perp }={\stackrel{\rightharpoonup }{v}}_2-\left({\stackrel{\rightharpoonup }{u}}_1·{\stackrel{\rightharpoonup }{v}}_2\right){\stackrel{\rightharpoonup }{u}}_1$:

$\begin{array}{rcl}{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }& =& {\stackrel{\rightharpoonup }{v}}_2-\left({\stackrel{\rightharpoonup }{u}}_1·{\stackrel{\rightharpoonup }{v}}_2\right){\stackrel{\rightharpoonup }{u}}_1\\ & =& \frac{1}{2}\left[\begin{array}{c}9\\ -1\\ -1\\ 9\end{array}\right]-\frac{4}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\\ & =& \frac{1}{2}\left[\begin{array}{c}9\\ -1\\ -1\\ 9\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right]\\ & =& \frac{1}{2}\left[\begin{array}{c}5\\ -5\\ -5\\ 5\end{array}\right]\end{array}$

So, the vector u₂ is:

$\begin{array}{rcl}{u}_2& =& \frac{{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }}{\left|\left|{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }\right|\right|}\\ & =& \frac{1}{5}\left(\frac{1}{2}\left[\begin{array}{c}5\\ -5\\ -5\\ 5\end{array}\right]\right)\\ & =& \frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\end{array}$

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" $R=\left[\begin{array}{cc}\left|\left|v_1\right|\right|& {\stackrel{\rightharpoonup }{u}}_1·{\stackrel{\rightharpoonup }{v}}_2\\ 0& \left|\left|{{\stackrel{\rightharpoonup }{v}}_2}^{\perp }\right|\right|\end{array}\right]$ as follows:

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

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 $\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& -1\\ 1& -1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 4\\ 0& 5\end{array}\right]$.