Science:Math Exam Resources/Courses/MATH152/April 2022/Question A14

MATH152 April 2022

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Question A14
Find a matrix ${\textstyle A}$ that has eigenvectors ${\textstyle {\vec {v}}_{1}=\left[{\begin{array}{c}1\\0\end{array}}\right]}$ and ${\textstyle {\vec {v}}_{2}=\left[{\begin{array}{c}2\\1\end{array}}\right]}$, corresponding to eigenvalues ${\textstyle \lambda _{1}=1}$ and ${\textstyle \lambda _{2}=2}$.

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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
We know that for a matrix ${\textstyle A}$, the first column is ${\textstyle A{\vec {e}}_{1}}$, and the second column is ${\textstyle A{\vec {e}}_{2}}$. How can you relate the standard basis vectors ${\textstyle {\vec {e}}_{1}}$ and ${\textstyle {\vec {e}}_{2}}$ to the given eigenvectors?

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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. Recall from the hint that a matrix ${\textstyle A}$ will have first column ${\textstyle A{\vec {e}}_{1}}$, and second column ${\textstyle A{\vec {e}}_{2}}$, where ${\textstyle {\vec {e}}_{1}}$ and ${\textstyle {\vec {e}}_{2}}$ are the standard basis vectors ${\textstyle [1\ \ 0]^{T}}$ and ${\textstyle [0\ \ 1]^{T}}$. Now, note that ${\textstyle {\vec {v}}_{1}={\vec {e}}_{1}}$, and use the property that ${\textstyle A{\vec {v}}=\lambda {\vec {v}}}$ for an eigenvalue-eigenvector pair to get $$A{\vec {e}}_{1}=A{\vec {v}}_{1}=\lambda _{1}{\vec {v}}_{1}=\lambda _{1}{\vec {e}}_{1}={\vec {e}}_{1},$$ since ${\textstyle \lambda _{1}=1}$. This results in the first column of ${\textstyle A}$ simply being ${\textstyle {\vec {e}}_{1}}$. For the second column of ${\textstyle A}$, we first note that ${\textstyle {\vec {e}}_{2}={\vec {v}}_{2}-2{\vec {v}}_{1}}$. Using the same eigenvalue-eigenvector relation we have $$A{\vec {e}}_{2}=A({\vec {v}}_{2}-2{\vec {v}}_{1})=A{\vec {v}}_{2}-2A{\vec {v}}1=\lambda _{2}{\vec {v}}_{2}-2\lambda _{1}{\vec {v}}_{1}=2{\vec {v}}_{2}-2{\vec {v}}_{1},$$ since ${\textstyle \lambda _{2}=2}$. This results in the second column being ${\textstyle [2\ \ 2]^{T}}$, and the matrix is $$A=\left[{\begin{array}{cc}1&2\\0&2\end{array}}\right].$$