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Let T be a linear operator on a finite dimensional vector space $V$, and suppose that $W$ is a T-invariant subspace of V. Prove that the minimal polynomial of $T_w$ divides the minimal polynomial of T.

I tried this:

Let $p_1(t)$ be the minimal polynomial of $T$. Let $p_2(t)$ be the minimal polynomial of $T_w$.

By the division algorithm for polynomials, there exist polynomials q(t) and r(t) such that

$$p_1(t)=p_2(t)q(t)+r(t).$$

where r(t) has degree less than $p_2(t)$.

If we choose $w \in W$

$$p_1(T)(w)=p_2(T)(w)q(T)(w)+r(T)(w).$$

then

$$r(T)(w)=0$$

This by hypothesis.

Hence, $$p_1(t)=p_2(t)q(t).$$

And, $$p_2(t)|p_1(t).$$

Is this correct?

Bayesian guy
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  • Sorry, how $r(T)(w) = 0 \implies r(t)=0$? – azif00 Jun 23 '21 at 20:10
  • This is the step I think is wrong but I don't know how I can prove the statement. – Bayesian guy Jun 23 '21 at 22:17
  • Usually when reasoning about the remainder in a Euclidean division (especially when trying to prove it is zero), one needs to use the degree bound (strictly less than the degree of the divisor), since without this it need not be the actual remainder. Your proof never invokes the degree bound, which is a red flag. – Marc van Leeuwen Jun 29 '21 at 11:54
  • Thanks a lot for the tip, I already did the problem. I think I need more advices like yours. – Bayesian guy Jul 05 '21 at 16:20

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