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

MATH152 April 2015

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Question B 6 (a)
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. (a) Find all the eigenvalues of $A$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Solve the equation ${\text{det}}(A-\lambda I)=0$.

Hint 2
We may also use the property of orthogonal rotation matrix as the matrix $A$ represents a rotation.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. $A-\lambda I={\begin{bmatrix}1/2-\lambda &-{\sqrt {2}}/2&1/2\\{\sqrt {2}}/2&-\lambda &-{\sqrt {2}}/2\\1/2&{\sqrt {2}}/2&1/2-\lambda \end{bmatrix}}$ Remember when we calculate the determinant: suppose we have a 3 × 3 matrix A, and we want the specific formula for its determinant : $|A|={\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}=a\,{\begin{vmatrix}e&f\\h&i\end{vmatrix}}-b\,{\begin{vmatrix}d&f\\g&i\end{vmatrix}}+c\,{\begin{vmatrix}d&e\\g&h\end{vmatrix}}$ Back to our problem, ${\begin{aligned}|A-\lambda I|&=(1/2-\lambda )[(-\lambda )(1/2-\lambda )-(-{\sqrt {2}}/2)({\sqrt {2}}/2)]-(-{\sqrt {2}}/2)[({\sqrt {2}}/2)(1/2-\lambda )-(-{\sqrt {2}}/2)1/2]+1/2[({\sqrt {2}}/2)({\sqrt {2}}/2)-1/2(-\lambda )]\\&=-(\lambda -1)(\lambda ^{2}+1)=0,\end{aligned}}$ so that $\lambda _{1}=1,\lambda _{2}=i,\lambda _{3}=-i$ answer: $\color {blue}{1,i,-i}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the second hint, Since matrix $A$ is a orthogonal rotation matrix, therefore it has determinant $\pm 1$. For this question, the determinant is $1$. Out of three eigenvalues, one of them must be 1 from the property of an orthogonal rotation matrix. Now suppose the left two eigenvalues are $z_{1}$ and $z_{2}$. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have $1+z_{1}+z_{2}=1$ and $z_{1}\cdot z_{2}=1$. $z_{1}=i,z_{2}=-i$ answer: $\color {blue}1,i,-i$