Science:Math Exam Resources/Courses/MATH152/April 2022/Question B3 (a)

MATH152 April 2022

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Question B3 (a)
Let $$A=\left[{\begin{array}{ccc}0&5&-5\\1&4&1\\0&0&5\end{array}}\right].$$ (a) Find the eigenvalues of $A$.

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
Recall the definition: an eigenvalue of $A$ is a number $\lambda$ such that the (vector) equation $$Av=\lambda v$$ has a solution. Can you rewrite the above equation so that the vector $v$ is in the null-space of a matrix?

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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. Here is the argument for the standard procedure that computes the eigenvalues of a matrix. The equation $$Av=\lambda v$$ is equivalent to $$(A-\lambda \mathrm {Id} )v=0.$$ This latter equation has a non-zero solution if and only if the matrix $A-\lambda \mathrm {Id}$ is not invertible, which is equivalent to $\det(A-\lambda \mathrm {Id} )=0$. Therefore, we solve this last equation, since it is an equation of numbers (as opposed to an equation of vectors). We find the determinant to be $${\begin{aligned}\det(A-\lambda \mathrm {Id} )&=\det \left[{\begin{array}{ccc}-\lambda &5&-5\\1&4-\lambda &1\\0&0&5-\lambda \end{array}}\right]\\&=(5-\lambda )\det \left[{\begin{array}{cc}-\lambda &5\\1&4-\lambda \end{array}}\right]\\&=(5-\lambda )(-\lambda (4-\lambda )-5)=(5-\lambda )(\lambda ^{2}-4\lambda -5)=(5-\lambda )(\lambda -5)(\lambda +1).\end{aligned}}$$ From the factorization above, we see that there are two eigenvalues of $A$: -1 and 5, and 5 is called a repeated eigenvalue, since it shows up more than once in the factorization of $\det(A-\lambda \mathrm {Id} )$.