Finding linear transform matrix from characteristic polynomial I got two similiar very simple question on a notebook. 
1)let characteristic polynomial $P_A(x)=x^2+2x-3$
and
$T:V\to V$ and DimV=2,S={$\alpha_1,\alpha_2$} is base
find the matrix $[T]_S=A$
2)$P_A(x)=x^2+5x-7$
V,       $T:V \to V$, S={$\alpha_1,\alpha_2$}
find the matrix $[T]_S=A$
there are solution under the questions which I didnt understand.
in first:
$T(\alpha_1)=\alpha_2=0\alpha_1+\alpha_2$
$T(\alpha_2)=-3\alpha_1-2\alpha_2$
$[T]_S=\begin{pmatrix} 
0 &  -3 \\ 
1 &  -2
\end{pmatrix}=A$
question: how did he wrote $T(\alpha_1)$ and $T(\alpha_2)$ equation? I know $T(\alpha_1)=\alpha_1$ and $T(\alpha_2)=-3\alpha_2$ since 1 and -3 are eigen values.
and shouldnt matrix be wrong? because $|xI-A|=\begin{vmatrix} 
x &  3 \\ 
-1 &  x+2
\end{vmatrix}=x^2+2x+3$ 
which doesnt even have root.
in second question, 
solution was 
$T(\alpha_1)=\alpha_2$,
$T(\alpha_2)=7\alpha_1-5\alpha_2$
$[T]_S=\begin{pmatrix} 
0 &  7 \\ 
1 &  -5
\end{pmatrix}=A$ 
question: again the equation of $T(\alpha_1),T(\alpha_2)$

For a $2 \times 2$ matrix $A$, you can check that
$$
p_A(x) = x^2 - ( \operatorname{tr}A ) x + ( \det A ).
$$
So, if
$$
p_A(x) = x^2 - T x + D,
$$
for some numbers, $T$ and $D$, you have to come up with the $2^2 = 4$ entries of the matrix satisfying
$$
\left\{
\begin{align}
a + d &= T \\
ad - bc &= D
\end{align}
\right.
$$
The solution is not unique.  In fact, there's a $2$-dimensional surface worth of solutions.
