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If I have an upper triangle matrix $$ \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

I do understand that the characteristic polynomial is $(x-1)^3$. But can I simply conclude that the minimal polynomial is $(x-1)$ in this case? Thank you!

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  • $\begingroup$ no, in fact you know it isn't, since $A-I\ne 0$ $\endgroup$ – qbert Apr 21 '18 at 3:25
  • $\begingroup$ calculate $(A-I)^2$ $\endgroup$ – Will Jagy Apr 21 '18 at 3:36
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The minimal polynomial is the monic polynomial of least degree that kills your matrix (meaning plug your matrix into the polynomial in the obvious way and get back the zero matrix). It also divides any polynomial that kills the matrix.

It is also true (by Cayley Hamilton ) that the characteristic polynomial kills the matrix, meaning for you $(A-I)^3=0$. That leaves you to check the divisors of this polynomial for the minimal polynomial. By direct computation, $A-I$ and $(A-I)^2$ won't fly, but $(A-I)^3=0$. This means your minimal polynomial is your characteristic polynomial, $(x-1)^3$.

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  • 3
    $\begingroup$ Algebraists are so violent. $\endgroup$ – Misha Lavrov Apr 21 '18 at 3:40

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