Science:Math Exam Resources/Courses/MATH307/December 2008/Question 04/Solution 1

To begin with, we note that the columns of A are already orthogonal,

${\begin{bmatrix}0\\1\\0\\-1\\0\end{bmatrix}}\cdot {\begin{bmatrix}1\\0\\-1\\0\\1\end{bmatrix}}=0.$

So we can simply put the normalized columns of A into the matrix Q:

$Q={\begin{bmatrix}0&{\frac {1}{\sqrt {3}}}\\{\frac {1}{\sqrt {2}}}&0\\0&-{\frac {1}{\sqrt {3}}}\\-{\frac {1}{\sqrt {2}}}&0\\0&{\frac {1}{\sqrt {3}}}\end{bmatrix}}.$

As a last step, we calculate R = Q^TA to find

$R={\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {3}}\end{bmatrix}}.$

(Note that the question asks for the QR decomposition, not a QR decomposition. This is the unique QR decomposition of A such that Q has orthonormal columns and R is square upper triangular with positive diagonal entries.)