Science:Math Exam Resources/Courses/MATH152/April 2010/Question A 05
• 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 3(d) • QB 4(a) • QB 4(b) • QB 4(c) • QB 4(d) • QB 4(e) • QB 5(a) • QB 5(b) • QB 6(a) • QB 6(b) • QB 6(c) • QB 6(d) • QB 6(e) •
Question A 05 

If 2 and 3 are both eigenvalues of a matrix A then A must be invertible. Justify this statement briefly. 
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 

How many eigenvalues does an matrix have. How do the eigenvalues relate to the invertibility of a matrix? 
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. Recall that matrices have n eigenvalues. Here since n=2 we have 2 eigenvalues which are given as 2 and 3. Also recall that if A is not invertible then there exists at least one nonzero vector x such that Ax=0. Therefore, we see that if A is not invertible then there is at least one zero eigenvalue with eigenvector x. Conversely, if 0 is an eigenvalue then there exists a nonzero vector x such that Ax=0 which means A is not invertible. This concludes that A being noninvertible implies a zero eigenvalue and also that a zero eigenvalue implies that A is noninvertible. We just wrote that the only eigenvalues are 2 and 3 so therefore, 0 is not an eigenvalues which means A must be invertible. 