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Let $V$ be a finite dimension vector space over $\mathbb{C}$. Let $T:V\to V$ be a linear transformation and suppose the minimal polynomial of $T$ is

$$x^{13}(x-1)$$

What is the minimal polynomial of $T^4$?

I know that the minimal polynomial of $T^4$ divides the minimal polynomial of $T$, as

$$\left(T^{4}\right)^{13}\left(T^{4}-{\rm Id}\right)=\left(T^{13}\right)\left(T-1\right)\cdot\left(T^{13}\right)^{3}\left(T^{3}+T^{2}+T+{\rm Id}\right)=0$$

this means it doesn't have any other roots. Similarly it's easy to check that both $1$ and $0$ are eigenvalues of $T^4$ as well, thus these are the only roots, but they may be of smaller order.

For $(x-1)$ it's obvious. But I have no idea how to find the degree of $x$ in the minimal polynomial.

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  • $\begingroup$ I would like to know why you say the minimal polynomial of $T^4$ divides the minimal polynomial of $T$. It happens to be the case here, but this kind of thing is not true for arbitrary $T$. $\endgroup$ Nov 19 '15 at 15:38
  • $\begingroup$ @MarcvanLeeuwen well, I give my reasoning in the question. Setting $T^4$ in the minimal polynomial of $T$ gives you zero, so the minimal polynomial of $T^4$ must divide it... $\endgroup$
    – Nescio
    Nov 19 '15 at 15:41
  • $\begingroup$ Fair enough. I'll give an answer, but that divisibility plays no role in it. $\endgroup$ Nov 19 '15 at 15:50
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Since that is the minimal polynomial of $T$, we have the relation $T^{14}=T^{13}$. With this, we can try to find a minimal polynomial for $T^4$. The first power where we get a power at least $14$ is $(T^4)^4=T^{16}$, which is $T^{13}$ by the relation. This does not get a linear dependence, so we go to the next power, $(T^4)^5=T^{20}=T^{13}$, so $(T^4)^5-(T^4)^4=0$, giving the minimal polynomial $x^4(x-1)$ for $T^4$.

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  • $\begingroup$ Unfortunately I did not understand the answer. How from $(T^4)^5-(T^4)^4=0$ did you get to the minimal polynomial? $\endgroup$
    – Nescio
    Nov 19 '15 at 8:23
  • $\begingroup$ Also, I'm not quite sure about the relation $T^14=T^13$. This is obviously true if you look only on the generalized eigenspace of $0$, but what makes it hold true for all of $V$? $\endgroup$
    – Nescio
    Nov 19 '15 at 8:25
  • $\begingroup$ Well took me a while, but I understand everything. Thank you $\endgroup$
    – Nescio
    Nov 19 '15 at 9:42
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Given that minimal polynomial, with $x^{13}$ and $x-1$ relatively prime, the kernel decomposition theorem says your space decomposes as a direct sum of $\ker(T^{13})$ (the generalised eigenspace for eigenvalue$~0$) and $\ker(T-I)$ (the eigenspace for eigenvalue$~1$). A polynomial $P[T^4]$ of $T^4$ will be zero iff it acts as zero on both summands, so the minimal polynomial of $T^4$ is the least common multiple of the minimal polynomials of the restrictions of $T^4$ to those summands.

On the summand $\ker(T-I)$, $T^4$ acts as multiplication by $1^4=1$, so its minimal polynomial is $X-1$. On the other summand $\ker(T^{13})$, the restriction of $T$ has minimal polynomial $x^{13}$. But then it is easy to see the minimal polynomial of the restriction of $T^4$ to this summand is $x^4$ (since $x^4[T^4]$ is zero on the summand, but not $Q[T^4]$ for any strict divisor $Q$ of $x^4$, notably not for $Q=x^3$). All in all the minimal polynomial of$~T^4$ is $$ \operatorname{lcm}(x-1,x^4)=(x-1)x^4. $$

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