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 2x1 matrix, determinant does not exist here, so how to find the matrix A as stated in the question?
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
$$A\left(
\begin{array}{c}
1 \\
1 \\
\end{array}
\right)=\frac{2}{3}A\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)-\frac{1}{3}A\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)=....\mbox{(using $(1)$)}$$
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