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

MATH152 April 2022

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[hide] 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 $A$.

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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!

[show] 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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[show] Solution

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 $\det(A-\lambda \mathrm {Id} )$.