Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (b)

MATH152 April 2015

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Question B 6 (b)
The matrix $A={\begin{bmatrix}1/2&-{\sqrt {2}}/2&1/2\\{\sqrt {2}}/2&0&-{\sqrt {2}}/2\\1/2&{\sqrt {2}}/2&1/2\end{bmatrix}}$ represents a rotation in 3D relative to some axis. (b) Find the direction vector of the axis of the rotation. Hint: this vector remains unchanged after the rotation.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
If a vector is unchanged by a linear transformation, that means it is an eigenvector with eigenvalue 1.

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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. We are looking for the eigenvector ${\textstyle v}$ associated with eigenvalue ${\textstyle \lambda =1}$. This means that ${\textstyle v}$ solves the equation $$(A-\lambda I){\mathbf {v}}={\begin{bmatrix}-1/2&-{\sqrt {2}}/2&1/2\\{\sqrt {2}}/2&-1&-{\sqrt {2}}/2\\1/2&{\sqrt {2}}/2&-1/2\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}=0$$ It remains to solve for ${\textstyle {\mathbf {v}}}$. We do this by row reduction, beginning with dividing the second row by ${\textstyle {\sqrt {2}}}$: $${\begin{bmatrix}-1/2&-{\sqrt {2}}/2&1/2\\1/2&-{\sqrt {2}}/2&-1/2\\1/2&{\sqrt {2}}/2&-1/2\end{bmatrix}}$$ Adding the first row to the second and third rows: $${\begin{bmatrix}-1/2&-{\sqrt {2}}/2&1/2\\0&-{\sqrt {2}}&0\\0&0&0\end{bmatrix}}$$ Lastly we add ${\textstyle (-1/2)}$ times the second row to the first row and multiply the first row by ${\textstyle -2}$ and the second row by ${\textstyle -{\frac {1}{\sqrt {2}}}}$ (to get leading zeros): $${\begin{bmatrix}1&0&-1\\0&1&0\\0&0&0\end{bmatrix}}$$ Translating this back into a system of equations, this matrix is saying ${\textstyle v_{1}-v_{3}=0}$ and ${\textstyle v_{2}=0}$. Thus, this operation rotates vectors around the axis spanned by the vector $$\color {blue}{\begin{bmatrix}1\\0\\1\end{bmatrix}}$$