Science:Math Exam Resources/Courses/MATH221/December 2011/Question 04 (b)

MATH221 December 2011

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Question 04 (b)
True or false (explain your answer): If v is an eigenvector for the invertible matrix A, then v is also an eigenvector for the matrix A−1.

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
What does it mean to be an eigenvector v of a matrix A? Write down the definition and start from there.

Hint 2
Since we are told that A is invertible, we know that A-1 exists. So you can multiply that to your equation that defines an eigenvector v.

Hint 3
Can you divide by the eigenvalue λ? I.e. could λ be zero?

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Step 1. If v is an eigenvector for a matrix A for some eigenvalue λ, then we can write $\displaystyle Av=\lambda v$ Step 2. Since we assume that the matrix A is invertible, we can multiply on the above equation by that inverse on the left on each side and obtain $\displaystyle A^{-1}Av=A^{-1}\lambda v$ Which is equivalent to $\displaystyle v=\lambda A^{-1}v$ Step 3. We would now like to divide by λ, but before we should check that this is allowed, i.e. that λ is not zero. Since A in invertible, Aw = 0 = 0(w) implies w = 0. Thus λ = 0 is not an eigenvalue for A and we can safely bring λ to the LHS in the equation above. This yields $\displaystyle {\frac {1}{\lambda }}v=A^{-1}v$ which would mean that v is an eigenvector for the eigenvalue 1/λ of the inverse of the matrix A. So we can conclude that the statement is true.