Science:Math Exam Resources/Courses/MATH307/December 2010/Question 07 (a)

MATH307 December 2010

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Question 07 (a)

Suppose that $A$ is a symmetric $4\times 4$ matrix with eigenvalues 0, 1, 4, 5. Define a sequence of vectors $\mathbf {x} _{n}\in \mathbb {R} ^{4}$ by choosing $\mathbf {x} _{0}$ at random, and then settings ${\begin{aligned}\mathbf {y} _{n}&=(A-3I)^{-1}\mathbf {x} _{n-1}\\\mathbf {x} _{n}&=\mathbf {y} _{n}/\|\mathbf {y} _{n}\|\end{aligned}}$ for $n=1,2,\dots$. You then observe that $\mathbf {x}$n converges to $\mathbf {x} _{\infty }=[1/2,1/2,1/2,1/2]^{T}$ as $n\to \infty$. (a) What is $A\mathbf {x} _{\infty }$?

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
In the power iteration procedure, the convergent vector $\displaystyle x_{\infty }$ is the eigenvector corresponding to the largest eigenvalue of the matrix in question. In this question the matrix is $\displaystyle (A-3I)^{-1}$

Hint 2
The eigenvalues of $\displaystyle A-3I$ are the eigenvalues of A shifted by 3. For the inverse matrix, take the inverse of these eigenvalues. Find the eigenvalue of $\displaystyle (A-3I)^{-1}$ with the largest absolute value, as well as the corresponding eigenvalue of A.

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. Using the hints, we know that $\displaystyle \mathbf {x} _{\infty }$ is the eigenvector associated to the largest eigenvalue of the matrix $\displaystyle (A-3I)^{-1}$. The eigenvalues of $\displaystyle A-3I$ are the eigenvalues of $A$ shifted by 3, with the same eigenvectors. We can see this by noting for each eigenvector $v$ of $A$, we have $\displaystyle (A-3I)v=Av-3v=\lambda v-3v=(\lambda -3)v$ Since $A$ has the eigenvalues $\lambda _{0}=0$, $\lambda _{1}=1$, $\lambda _{2}=4$ and $\lambda _{3}=5$, the matrix $\displaystyle A-3I$ has the eigenvalues $\eta _{0}=-3$, $\eta _{1}=-2$, $\eta _{3}=1$ and $\eta _{4}=2$. Next, the eigenvalues of the inverse $\displaystyle (A-3I)^{-1}$ are the reciprocal of the original eigenvalues, with the same eigenvectors. Hence the eigenvalues of $\displaystyle (A-3I)^{-1}$ are $\gamma _{0}=-1/3$, $\gamma _{1}=-1/2$, $\gamma _{3}=1$ and $\gamma _{4}=1/2$. We see that the dominating eigenvalue of $\displaystyle (A-3I)^{-1}$ is $\gamma _{3}=1$, and therefore, by the power method, $\mathbf {x} _{\infty }$ is an eigenvector of $\displaystyle (A-3I)^{-1}$ with eigenvalue 1. This means that $\mathbf {x} _{\infty }$ is also an eigenvector of $A$ with eigenvalue $\lambda _{2}=4$. In other words, $\displaystyle A\mathbf {x} _{\infty }=4\mathbf {x} _{\infty }=[2,2,2,2]^{T}$

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