Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 05 (c)
• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QA 9 • QA 10 • QA 11 • QA 12 • QA 13 • QA 14 • QA 15 • QA 16 • QA 17 • QA 18 • QA 19 • QA 20 • QA 21 • QA 22 • QA 23 • QA 24 • QA 25 • QA 26 • QA 27 • QA 28 • QA 29 • QA 30 • QB 1(a) • QB 1(b) • QB 1(c) • QB 2(a) • QB 2(b) • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 4(d) • QB 5(a) • QB 5(b) • QB 5(c) • QB 6(a) • QB 6(b) • QB 6(c) • QB 6(d) • QB 6(e) •
Question B 05 (c) 

Consider the matrix below:
It is known that is an eigenvalue of A. Find eigenvectors associated with and . 
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 

What was the special relationship between and . Does this relationship hold true for their eigenvectors? 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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. In part (b) we found that the other two eigenvalues and were complex conjugates, . Therefore, the eigenvectors for and will also occur in complex conjugate pairs. It is then sufficient to just find one of the vectors and then conjugate it to find the other. We will solve the vector for . This means we are seeking a vector for which (AI)x=0.
If we denote the vector x as [a,b,c] then we need to solve which is equivalent to solving If we solve this we get a=(1+i)c and b=ic for arbitrary c. We can set c=1 without loss of generality to get that the eigenvector for is This means automatically that the eigenvector for is just the conjugate of the eigenvector for . Therefore, If you still have trouble computing eigenvectors, it is a worthwhile exercise to actually compute using the method above and comparing with this result. 