Science:Math Exam Resources/Courses/MATH221/December 2007/Question 07 (c)

MATH221 December 2007

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Question 07 (c)
Consider the matrix $A={\begin{pmatrix}0&-4&-6\\-1&0&-3\\1&2&5\end{pmatrix}}$. For each eigenvalue, find the dimension of the corresponding eigenspace.

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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

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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. From part (b), we know that the eigenvalues of A are 1 and 2. We also saw in part (b) that the eigenvalue 2 is a repeated root of the characteristic polynomial for A, while the eigenvalue 1 is not. By definition, the dimension of each respective eigenspace is at least 1. As well, recall that the dimension of each eigenspace is at most the multiplicity of the eigenvalue as a root of the characteristic polynomial. As the eigenvalue 1 is not a repeated root of the characteristic polynomial for A, the dimension of its eigenspace must be at least 1, and at most 1, i.e. it is equal to 1. The eigenvalue 2 is a twice repeated root of the characteristic polynomial for A; therefore the dimension of its eigenspace may be either 1 or 2. We will need to investigate further to determine which one it is. To calculate the dimension of the eigenspace corresponding to the eigenvalue 2, we want to determine $\dim {\text{null}}(A-2I)$, that is, the dimension of the kernel of A - 2I. We have $A-2I={\begin{pmatrix}-2&-4&-6\\-1&-2&-3\\1&2&3\end{pmatrix}}.$ All three rows of A - 2I are multiples of each other, so ${\text{rank}}(A-2I)=1$. Therefore, by the rank-nullity theorem, $\dim {\text{null}}(A-2I)=\dim(\mathbb {R} ^{3})-\dim {\text{rank}}(A-2I)=3-1=2.$ In conclusion, For the eigenvalue 1, the dimension of its eigenspace is 1. For the eigenvalue 2, the dimension of its eigenspace is 2.

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