Science:Math Exam Resources/Courses/MATH152/April 2022/Question A15

MATH152 April 2022

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Question A15
For which value of ${\textstyle b}$ does the matrix ${\textstyle A}$ below have two linearly independent eigenvectors corresponding to the eigenvalue ${\textstyle \lambda =1}$? $$A=\left[{\begin{array}{ccc}1&-4&b\\0&3&3\\0&0&1\end{array}}\right]$$

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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
If ${\textstyle A}$ has 2 linearly independent eigenvectors corresponding to ${\textstyle \lambda =1}$, then the null-space of $A-\lambda I$ must be 2-dimensional; equivalently, we must have ${\textstyle \mathbf {rank} (A-\lambda I)=1}$. Is there a simple way to pick ${\textstyle b}$ so that ${\textstyle \mathbf {rank} (A-\lambda I)=1}$?

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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. Since we want ${\textstyle A}$ to have 2 linearly independent eigenvectors associated with ${\textstyle \lambda =1}$, then we would have ${\textstyle \mathbf {rank} (A-\lambda I)=1}$. This means that using row operations, we should be able to reduce the matrix ${\textstyle A-\lambda I}$ to a single row! We need to find ${\textstyle b}$ such that this holds. $$A-\lambda I=\left[{\begin{array}{ccc}0&-4&b\\0&2&3\\0&0&0\end{array}}\right]\rightarrow \left[{\begin{array}{ccc}0&2&3\\0&0&b+6\\0&0&0\end{array}}\right].$$ Then, in order to have ${\textstyle A-\lambda I}$ reduce to a single row, we take ${\textstyle b=-6}$.