Find matrix from Eigenvectors and Eigenvalues 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?
 A: 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
$$A\left(  
\begin{array}{c}
    1 \\
    1 \\
  \end{array}
\right)=\frac{2}{3}A\left(  
\begin{array}{c}
    2 \\
    1 \\
  \end{array}
\right)-\frac{1}{3}A\left(  
\begin{array}{c}
    1 \\
    -1 \\
  \end{array}
\right)=....\mbox{(using $(1)$)}$$
A: Let us try without computing $A$. To do that we have to decompose b as a linear combination of $v_1$ and $v_2$ like $\mathbf{b}=\alpha_1 v_1+\alpha_2 v_2$ And this would yield $$A\mathbf{b}=\alpha_1\lambda_1 v_1+\alpha_2\lambda_2 v_2$$ To find $\alpha_1$ and $\alpha_2$ we just have to solve a set of two linear equations $$\begin{cases} 2\alpha_1+\alpha_2=1\\ \alpha_1-\alpha_2=1 \end{cases}$$ (solution $\alpha_1=\frac{2}{3}$ and $\alpha_2=\frac{-1}{3}$)
A: Note that 
$$
\vec b=\frac{2}{3}\,\vec v_1-\frac{1}{3}\,\vec v_2
$$
Hence
$$
A\vec b=\frac{2}{3}A\vec v_1-\frac{1}{3}A\vec v_2
$$
A: You don't have to know $A$ explicitly. All you need to know is that $Av_1=2v_1$ and $Av_2=-3v_2$. Since $b=\frac{2}{3}v_1-\frac{1}{3}v_2$ we have
\begin{align*}
Ab=A(2/3v_1-1/3v_2)=1/3(2Av_1-Av_2)=1/3(4v_1+3v_2)=1/3(11, 1).
\end{align*}
But if you want to determine $A$, note that $TAT^{-1}=D$, where $D$ is the diagonal matrix of the eigenvalues of $A$, and the columns of $T$ are the corresponding eigenvectors of $A$.
