Science:Math Exam Resources/Courses/MATH152/April 2022/Question A20/Solution 1

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

$$A^{4}=\left[{\begin{array}{cc}-4&0\\0&-4\end{array}}\right].$$