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Thread: polinomial squared problem

  1. #1
    MHF Contributor
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    polinomial squared problem

    4)A)
    there is $\displaystyle A=\left(\begin{array}{ccc}i & 0 & 1\\0 & 3 & 0\\0 & 0 & i\end{array}\right)$ find polinomial $\displaystyle 0\neq Q(t)=at^{2}+bt+c\in C_{3}[t]$
    so $\displaystyle [Q(A)]^{2}=0$
    B)there is $\displaystyle A\in M_{nxn}^{F}$ matrices which is not diagonisable.
    prove that if F=C (the field is C) then there is a polinomial $\displaystyle 0\neq Q(t)\in C_{n}[t] so [Q(A)]^{2}=0$
    c)
    does the previos part id true if F=R
    ?
    dont knot from where to start,any starting guidence?
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  2. #2
    MHF Contributor
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    Re: polinomial squared problem

    i solved part A by taking the caracteristic polinomial of A$\displaystyle p(t)=(t-i)^{2}(t-3)$
    then we take d(t)=(t-i)(t-3)

    $\displaystyle d^{2}(t)=(t-i)^{2}(t-3)^{2}=(t-i)^{2}(t-3)(t-3)$

    by cayly hemilton

    $\displaystyle p(A)=(A-iI)^{2}(A-3I)=0$

    so

    $\displaystyle d^{2}(A)=(A-iI)^{2}(A-3I)^{2}=(A-iI)^{2}(A-3I)(A-3I)=0(A-3I)=0$

    what is the difference between part A and part B

    why cant i solve it the same way?
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