Science:Math Exam Resources/Courses/MATH221/December 2008/Question 05 (a)

MATH221 December 2008

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Question 05 (a)
Find the eigenvalues and eigenvectors for $A={\begin{bmatrix}5&0\\2&1\end{bmatrix}}$.

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
Recall that if $(A-\lambda I)v=0$, then $\lambda$ is an eigenvalue of $A$ with eigenvector $v$.

Hint 2
First, calculate the eigenvalues $\lambda _{1}$ and $\lambda _{2}$. To find the corresponding eigenvectors, find vectors in the kernel of $(A-\lambda _{1}I)$ and $(A-\lambda _{2}I)$.

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. To calculate the eigenvalues of $A$ we need to find values of $\lambda$ such that $A-\lambda I$ has a non-trivial kernel. A good way to check this is to look for zeros of the determinant of the matrix above: $0={\text{det}}(A-\lambda I)={\text{det}}{\begin{bmatrix}5-\lambda &0\\2&1-\lambda \end{bmatrix}}=(5-\lambda )(1-\lambda )-2(0)=(5-\lambda )(1-\lambda ).$ Luckily for us, the polynomial is already factorized, so that we can simply read off the eigenvalue $\lambda _{1}=1$ and $\lambda _{2}=5$. To find the eigenvector $v={\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}$ corresponding to the eigenvalue $\lambda _{1}=1$, we solve $(A-I)v=0$. This equation is ${\begin{bmatrix}4&0\\2&0\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}},$ and thus $v={\begin{bmatrix}0\\1\end{bmatrix}}$ is an eigenvector associated to $\lambda _{1}=1$. Repeating the same steps for $\lambda _{2}=5$ gives us the eigenvector equation ${\begin{bmatrix}0&0\\2&-4\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}},$ and thus $v={\begin{bmatrix}2\\1\end{bmatrix}}$ is an eigenvector corresponding to $\lambda _{2}=5$. To summarize, $1$ is an eigenvalue of $A$ with corresponding eigenvector $v={\begin{bmatrix}0\\1\end{bmatrix}}$, and $5$ is an eigenvalue of $A$ with corresponding eigenvector ${\begin{bmatrix}2\\1\end{bmatrix}}$.

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