Science:Math Exam Resources/Courses/MATH152/April 2022/Question A20

MATH152 April 2022

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Question A20
Let ${\textstyle A}$ be a ${\textstyle 2\times 2}$ matrix with eigenvalues ${\textstyle 1+i}$, ${\textstyle 1-i}$. Find ${\textstyle A^{4}}$. (Hint: how does ${\textstyle A^{4}}$ transform the eigenvectors?)

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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
Looking at how ${\textstyle A^{4}}$ acts on the eigenvectors, you should be seeing something that looks like ${\textstyle A^{4}{\vec {v}}_{1}=c{\vec {v}}_{1}}$, ${\textstyle A^{4}{\vec {v}}_{2}=c{\vec {v}}_{2}}$, ${\textstyle c\in \mathbb {C} }$. Using this, and the fact that linearly independent eigenvectors span ${\textstyle \mathbb {R} ^{2}}$, what can you conclude about ${\textstyle A^{4}}$?

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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. We start by looking at how ${\textstyle A^{4}}$ transforms the eigenvectors. We don’t need to know what the eigenvectors are to do this, we just need the relation ${\textstyle A{\vec {v}}=\lambda {\vec {v}}}$. ${\textstyle \lambda _{1}=1+i}$: eigenvector is ${\textstyle {\vec {v}}_{1}}$, and ${\textstyle A^{4}{\vec {v}}_{1}=\lambda _{1}^{4}{\vec {v}}_{1}=(1+i)^{4}{\vec {v}}_{1}}$. Note that ${\textstyle 1+i={\sqrt {2}}e^{i{\frac {\pi }{4}}}}$, then ${\textstyle \lambda _{1}^{4}=4e^{i\pi }=-4}$, and ${\textstyle A^{4}{\vec {v}}_{1}=-4{\vec {v}}_{1}}$. ${\textstyle \lambda _{2}=1-i}$: eigenvector is ${\textstyle {\vec {v}}_{2}}$, and ${\textstyle A^{4}{\vec {v}}_{2}=(1-i)^{4}{\vec {v}}_{2}}$. Note ${\textstyle 1-i={\sqrt {2}}e^{-i{\frac {\pi }{4}}}}$, then ${\textstyle \lambda _{2}^{4}=4e^{-i\pi }=-4}$, and ${\textstyle A^{4}{\vec {v}}_{2}=-4{\vec {v}}_{2}}$. Because ${\textstyle {\vec {v}}_{1}}$ and ${\textstyle {\vec {v}}_{2}}$ are eigenvectors associated with distinct eigenvalues, we know they are linearly independent, and span ${\textstyle \mathbb {R} ^{2}}$. And, we have that ${\textstyle A^{4}}$ transforms ${\textstyle {\vec {v}}_{1}}$ and ${\textstyle {\vec {v}}_{2}}$ in the same way. So we must have that ${\textstyle A^{4}}$ is the diagonal matrix given by $$A^{4}=\left[{\begin{array}{cc}-4&0\\0&-4\end{array}}\right].$$