3) (15 points) The matrix $A=\left[\begin{array}{ll}3 & 1 \\ 2 & 2\end{array}\right]$ has eigenvectors \[ u=\left[\begin{array}{l} 1 \\ 1 \end{array}\right], \quad v=\left[\begin{array}{c} 1 \\ -2 \end{array}\right] \] a) (5 pts.) Determine the eigenvalue associated with $v$. b) (5 pts.) Write $x=\left[\begin{array}{l}1 \\ 7\end{array}\right]$ as a linear combination of $u$ and $v$. c) (5 pts.) The eigenvalue for $u$ is 4 . Use your result in (b) to determine $A^{3} x$. Answer: a) $A v=\left[\begin{array}{ll}3 & 1 \\ 2 & 2\end{array}\right]\left[\begin{array}{c}1 \\ -2\end{array}\right]=\left[\begin{array}{c}1 \\ -2\end{array}\right]=1 v$, so $\lambda=1$. Eigenvalue associated with $v$ is 1 . b) $x=\left[\begin{array}{l}1 \\ 7\end{array}\right]=3 u-2 v$ c) $\lambda_{1}=4, u=\left[\begin{array}{l}1 \\ 1\end{array}\right]$ and $\lambda_{2}=1, v=\left[\begin{array}{c}1 \\ -2\end{array}\right]$ AMS210.03 YanYu,yan2000@ams.sunysb.edu NAME STUDENT ID \#
Solution
Step 1 ：Given the matrix $A=\left[\begin{array}{ll}3 & 1 \\ 2 & 2\end{array}\right]$ and the eigenvector $v=\left[\begin{array}{c}1 \\ -2\end{array}\right]$
Step 2 ：We can find the eigenvalue associated with $v$ by multiplying the matrix $A$ with the eigenvector $v$. The result should be a scalar multiple of the eigenvector $v$, and this scalar is the eigenvalue.
Step 3 ：Performing the multiplication $A v=\left[\begin{array}{ll}3 & 1 \\ 2 & 2\end{array}\right]\left[\begin{array}{c}1 \\ -2\end{array}\right]=\left[\begin{array}{c}1 \\ -2\end{array}\right]$
Step 4 ：The result of the multiplication of the matrix $A$ and the eigenvector $v$ is the same as the eigenvector $v$, which means that the eigenvalue associated with $v$ is 1.
Step 5 ：Final Answer: The eigenvalue associated with $v$ is \(\boxed{1}\)
