A matrix $A$ has eigenvectors
$v_1 = \left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)$
$v_2 = \left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)$
with corresponding eigenvalues $\lambda_1$= 2 and $\lambda_2$= -3, respectively.
Determine Ab for the vector b = $
\left(
\begin{array}{c}
1 \\
1 \\
\end{array}
\right)$
I know how to find eigenvalues and eigenvectors from a given matrix A, but not this one,
the vector A is a 2×1 matrix, determinant does not exist here, so how to find the matrix A as stated in the question?
Best Answer
By definition of eigenvalue and eigenvector, we have $$\tag{1}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)=2\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)\mbox{ and }A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=-3\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right).$$ Now, since $$\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right),$$ we have $$A\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=....\mbox{(using $(1)$)}$$