Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 29

MATH152 April 2016

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Question A 29
The matrix below represents rotation in 3D about a line through the origin. $\left[{\begin{array}{ccc}1/2&-1/{\sqrt {2}}&1/2\\1/{\sqrt {2}}&0&-1/{\sqrt {2}}\\1/2&1/{\sqrt {2}}&1/2\end{array}}\right]$ Find a vector in the direction of the line of rotation.

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Hint
Consider the eigenvector. The direction vector of the rotation is just the vector $v$ unchanged after rotation, i.e., $Av=v.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If a vector lies in the line of rotation, after multiplying the rotational matrix on the vector, the vector stays the same. That is to say if we denote this vector by $v$, then we have $Av=v=1\cdot v$ Equivalently we are looking for eigenvector corresponding to eigenvalue 1. We need to solve characteristic equation $(A-I)v=0$ That is $${\begin{pmatrix}-{\frac {1}{2}}&-{\frac {1}{\sqrt {2}}}&{\frac {1}{2}}\\{\frac {1}{\sqrt {2}}}&-1&-{\frac {1}{\sqrt {2}}}\\{\frac {1}{2}}&{\frac {1}{\sqrt {2}}}&-{\frac {1}{2}}\end{pmatrix}}\cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}=0.$$ By row reduction we have $${\begin{pmatrix}-{\frac {1}{2}}&-{\frac {1}{\sqrt {2}}}&{\frac {1}{2}}\\{\frac {1}{\sqrt {2}}}&-1&-{\frac {1}{\sqrt {2}}}\\{\frac {1}{2}}&{\frac {1}{\sqrt {2}}}&-{\frac {1}{2}}\end{pmatrix}}\rightarrow {\begin{pmatrix}-{\frac {1}{2}}&-{\frac {1}{\sqrt {2}}}&{\frac {1}{2}}\\0&-2&0\\0&0&0\end{pmatrix}}\rightarrow {\begin{pmatrix}-1&-{\sqrt {2}}&1\\0&1&0\\0&0&0\end{pmatrix}}\rightarrow {\begin{pmatrix}-1&0&1\\0&1&0\\0&0&0\end{pmatrix}}.$$ Thus ${\textstyle -v_{1}+v_{3}=0}$ and ${\textstyle v_{2}=0}$, i.e., $${\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}={\begin{pmatrix}v_{1}\\0\\v_{1}\end{pmatrix}}=v_{1}{\begin{pmatrix}1\\0\\1\end{pmatrix}}.$$ So the eigenvector is $$\color {blue}(1,0,1)$$ which is the vector direction of axis.