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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

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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
$(AI)v=0$
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.

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