Science:Math Exam Resources/Courses/MATH221/December 2008/Question 10 (b)

MATH221 December 2008

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Question 10 (b)
True or false (if true, give a proof; if false, provide a counterexample): Suppose A is a 2 x 2 matrix with characteristic polynomial $(\lambda -2)^{2}$. Then A is diagonalizable.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A matrix is diagonalizable when it has distinct eigenvalues (why is true?).

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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. The statement is false. To construct a counterexample, we need to find a 2 x 2 matrix which has characteristic polynomial $(\lambda -2)^{2}$ but has only one linearly independent eigenvector. The first matrix that comes to mind which has characteristic polynomial $(\lambda -2)^{2}$ is the scalar multiple of the identity matrix, ${\begin{pmatrix}2&0\\0&2\\\end{pmatrix}}.$ However this clearly isn't a counterexample because it is a diagonal matrix. Can we modify it so that it isn't a diagonal matrix, but still has the same characteristic polynomial? A first simple modification is to investigate the diagonalizability of $A={\begin{pmatrix}2&1\\0&2\\\end{pmatrix}}.$ Our requirement that A has only one linearly independent eigenvector means there can be only one linearly independent solution to ${\begin{pmatrix}0&1\\0&0\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}},$ which is clearly the case, as ${\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}1\\0\end{pmatrix}}$ is the only linearly independent eigenvector. Therefore, ${\begin{pmatrix}2&1\\0&2\\\end{pmatrix}}$ is a counterexample to the statement.

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