Science:Math Exam Resources/Courses/MATH152/April 2022/Question A14/Solution 1

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

$$A=\left[{\begin{array}{cc}1&2\\0&2\end{array}}\right].$$