Find a 2x2 Matrix A for which: $E_{1}$ = span $\begin{bmatrix} 1 \\ 2 \end{bmatrix} $ and $E_{2}$ = $\begin{bmatrix} 2 \\ 3 \end{bmatrix} $ I am confused on how to approach this question. I am thinking of going backwards from the eigenvectors and and getting the matrix for which these eigenvectors came from, but having trouble doing that. Is this the right way to go about it? 
 A: Since $A$ is 2x2 and has two different eigenvalues, $\lambda_1=1$ and $\lambda_2=2$, they both have multiplicity one each. Therefore, $A$ is diagonalizable. Thus, we can write
$$
A=PDP^{-1}
$$
where
$$
D=
\begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix}
$$
and $P$ is the 2x2 matrix with the corresponding eigenvectors as columns. There aren't any unknowns. It should be a straight forward calculation.
A: So I found the solution for this question, but it doesn't make sense to me. It is solved by the following:
We want $A$ such that $A$ $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ = $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $A$ $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ = $2$$\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ = $\begin{bmatrix} 4 \\ 6 \end{bmatrix}$, i.e., $A$ = $\begin{bmatrix} 1 \ 4 \\ 2 \ 6 \end{bmatrix}$ (putting the two vectos into one matrix).
So $A$ $A^{-1}$ = $\begin{bmatrix} 5 \ -2 \\ 6 \ -2 \end{bmatrix}$.
How did they figure to do this with the eigenvectors? Is it what the user said above about the $E_{position}$ corresponding to the eigenvalue?    
