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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.

My try: Let T be a linear operator on a finite-dimensional vector space V, and suppose that W is a T-invariant subspace of V. let p(t) be the minimal polynomial of T. For $w \in W$, we have $p(T_w)(w)=P(T)(w)=0$ by definition of minimal polynomial. Hence proved.

Quotient space: Let V be any vector space and $W \subset V$ and buspace. For any $v in V$, let $v+W$ and denote the set : $v+W=\{v+w\mid w \in W\} \subset V $, called the coset of W containing v. Let $V/W=\{v+W\mid v \in V\}$ which we call the quotient space of $V$ modulo $W$.

Let T be a linear operator on a finite-dimensional vector space V, and suppose W is a T-invariant subspace of V. Let $\bar{T}:V/W \rightarrow V/W$ be a linear operator. Prove that the minimal polynomial of $\bar{T}$ divides the minimal polynomial of T.

My try: Let $p(\bar{T})$ be a polynomial of $\bar{T}$, then for any $v \in V$, $(g(\bar{T}))(v+W)=g(T)(v)+W=0+W=W$. Since it is T-invariant, by theorem( Let T be a linear operator on a finite-dimensional vector space V, and suppose that W is a T-invariant subspace of V, then the characteristic polynomial of $T_W$ divides the characteristic polynomial of T), we have proved the statement.

Is this right?

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    $\begingroup$ If you have shown that the minimal polynomial of $U$ divides any polynomial $g(t)$ such that $g(U)$ is the zero map, then yes. But it would be good to invoke this result explicitly, rather than elide it as you do in both cases. $\endgroup$ Apr 20, 2020 at 21:05

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