Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 29/Solution 1

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.