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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 $[T]_S=A$


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:



$[T]_S=\begin{pmatrix} 0 & -3 \\ 1 & -2 \end{pmatrix}=A$

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



$[T]_S=\begin{pmatrix} 0 & 7 \\ 1 & -5 \end{pmatrix}=A$

question: again the equation of $T(\alpha_1),T(\alpha_2)$

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up vote 2 down vote accepted

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.

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thanks. I know sol. is not unique but at the second quest.'s $T(\alpha_1)$ part. we know this from a+d=5, and ad-bc=-7 right? so can it be written as $T(\alpha_1)=-3\alpha_1+\alpha_2$ and $T(\alpha_2)=13\alpha_1-2\alpha_2$ if we take a=3,b=13,c=1,d=2 – lyme May 28 '14 at 22:22
Yeah. That works. – Sammy Black May 28 '14 at 22:36

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