Science:Math Exam Resources/Courses/MATH221/December 2011/Question 03 (a)

MATH221 December 2011

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Question 03 (a)
Consider the matrix $A=\left[{\begin{array}{cc}-6&8\\-4&6\end{array}}\right]$ Compute the eigenvalues of $A$ and a non-zero eigenvector for each eigenvalue.

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
Being clear with the definition of an eigenvalue and an eigenvector is really the key here.

Hint 2
Eigenvalues are values of the variable λ such that the system $\displaystyle A-\lambda I$ has a non-trivial nullspace.

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. Eigenvalues are values of the variable λ such that the system $\displaystyle A-\lambda I$ has a non-trivial nullspace or equivalently, when it has determinant 0. We compute this determinant: ${\begin{aligned}\det(A-\lambda I)&=\left|{\begin{array}{cc}-6-\lambda &8\\-4&6-\lambda \end{array}}\right|\\&=(-6-\lambda )(6-\lambda )-8(-4)\\&=-36+\lambda ^{2}+32\\&=\lambda ^{2}-4\end{aligned}}$ And so we conclude that this determinant is zero if and only if λ is 2 or -2. (Note: this equation is called the characteristic polynomial of the linear system). For each eigenvalue, an eigenvector is simply a non-trivial solution (non-trivial means not zero since that's an obvious solution that is always present). For λ = 2 we obtain the system ${\begin{aligned}A-2I&=\left[{\begin{array}{cc}-8&8\\-4&4\end{array}}\right]\\&\sim \left[{\begin{array}{cc}1&-1\\0&0\end{array}}\right]\\\end{aligned}}$ And so the general solution is a vector of the form [t, t ], which means that the eigenspace for the eigenvalue 2 is spanned by the vector [1, 1]. For λ = -2 we obtain the system ${\begin{aligned}A+2I&=\left[{\begin{array}{cc}-4&8\\-4&8\end{array}}\right]\\&\sim \left[{\begin{array}{cc}1&-2\\0&0\end{array}}\right]\\\end{aligned}}$ And so the general solution is a vector of the form [2t, t ], which means that the eigenspace for the eigenvalue -2 is spanned by the vector [2, 1].