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

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MATH221 December 2011

• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) •

Other MATH221 Exams

• December 2008 • December 2009 • December 2007 • April 2010 • April 2009 • December 2011 • April 2013 •

Question 04 (a)
Let $A=\left[{\begin{array}{ccc}2&0&0\\1&0&-1\\-1&2&3\end{array}}\right]$ Then is A diagonalizable? Explain your answer.

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 is a condition for a matrix to be diagonalizable?

Hint 2
You need to show that one can find three linearly independent eigenvectors.

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Solution
A matrix is diagonalizable if you can find a basis of eigenvectors, here that means we look for three linearly independent eigenvectors. To find eigenvectors, we first need to find eigenvalues. For this, we compute the characteristic polynomial: ${\begin{aligned}\det(A-\lambda I)&=\left\|{\begin{array}{ccc}2-\lambda &0&0\\1&-\lambda &-1\\-1&2&3-\lambda \end{array}}\right\|\\&=(2-\lambda )((-\lambda )(3-\lambda )+2)\\&=(2-\lambda )(\lambda ^{2}-3\lambda +2)\\&=(2-\lambda )(\lambda -2)(\lambda -1)\\&=-(\lambda -2)^{2}(\lambda -1)\end{aligned}}$ And so A has the eigenvalues 1 and 2. We now employ two facts about eigenvectors: The eigenvectors of distinct eigenvalues are linearly independent. The number of linearly independent eigenvector for a given eigenvalue (= dimension of the eigenspace = geometric multiplicity) is between 1 and the algebraic multiplicity of that eigenvalue. Recall that the algebraic multiplicity corresponds to the exponent in the characteristic polynomial: $\det(A-\lambda I)=-(\lambda -2)^{\color {red}2}(\lambda -1)^{\color {blue}1}$ means that the algebraic multiplicity of the eigenvalue 2 is two, and of the eigenvalue 1 is one. With this we conclude that There is exactly one eigenvector corresponding to the eigenvalue 1, and it is linearly independent from any eigenvector corresponding to the eigenvalue 2. There is either one or two eigenvectors corresponding to the eigenvalue 2. We can find out how many by calculating the dimension of the eigenspace. If the dimension of the eigenspace corresponding to the eigenvector 2 is two, then there are three linearly independent eigenvectors and A is diagonalizable. If the dimension is only one, then A has only two linearly independent eigenvectors and is not diagonalizable. Hence, we only need to find the dimension of that eigenspace. Since that eigenspace is the nullspace of the matrix A - 2I, we simply compute the rank of that matrix: ${\begin{aligned}A-2I&=\left[{\begin{array}{ccc}0&0&0\\1&-2&-1\\-1&2&1\end{array}}\right]\\&\sim \left[{\begin{array}{ccc}1&-2&-1\\0&0&0\\0&0&0\end{array}}\right]\\\end{aligned}}$ And since the rank is 1 this eigenspace will be of dimension 2. Therefore matrix A is diagonalizable.