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Let $W$ be an invariant subspace for $T$. Prove that the minimal polynomial for the restriction operator $T_W$ divides the minimal polynomial for $T$, without referring to matrices.

My attempt: We show $m_T(T_W)=0$. Let $x\in W$. Let $m_T=x^k+\sum_{i=0}^{k-1}a_i\cdot x^i$ be minimal polynomial of $T$. Then $$\begin{align}[m_T(T_W)](x) &= T_W^k(x)+\sum_{i=0}^{k-1}a_i\cdot T_W^i(x)\\ &= T^k(x)+\sum_{i=0}^{k-1}a_i\cdot T^i(x)\\ &=[m_T(T)](x)\\ &=0(x)\\ &=0. \end{align}$$ Since $x$ was arbitrary, we have $[m_T(T_W)](x)=0$, $\forall x\in W$. Thus $m_T(T_W)=0$. So $\exists q\in F[x]$ such that $m_T=m_{T_W}q$. Hence $m_{T_W}|m_T$. Is my proof correct?


Hoffman’s proof: We have $A=\begin{bmatrix} P& Q\\ 0& R\\ \end{bmatrix}$, where $A=[T]_B$ and $P=[T_W]_{B’}$. The $k$th power of the matrix $A$ has the block form $\begin{bmatrix} P^k& Q_k\\ 0& R^k\\ \end{bmatrix}$, where $Q_k$ is some $r\times (n-r)$ matrix. Let $f=\sum_{i=0}^nc_i\cdot x^i\in F[x]$ such that $f(A)=0$. Then $$f(A)= \sum_{i=0}^nc_i\cdot A^i= \sum_{i=0}^nc_i\cdot \begin{bmatrix} P^i& Q_i\\ 0& R^i\\ \end{bmatrix}= \sum_{i=0}^n \begin{bmatrix} c_i \cdot P^i& c_i\cdot Q_i\\ 0& c_i\cdot R^i\\ \end{bmatrix}= \begin{bmatrix} \sum_{i=0}^n c_i \cdot P^i & \sum_{i=0}^n c_i\cdot Q_i\\ 0& \sum_{i=0}^n c_i\cdot R^i\\ \end{bmatrix}= \begin{bmatrix}f(P) & \sum_{i=0}^n c_i\cdot Q_i \\ 0&f(R) \\ \end{bmatrix}= \begin{bmatrix} 0 &0 \\ 0&0 \\ \end{bmatrix}.$$ So $f(P)=0$. Let $m_T$ and $m_{T_W}$ be minimal polynomial of $T$ and $T_W$, respectively. Since $m_T(A)=0$ (by this post) , we have $m_T(P)=0$. So $m_T(T_W)=0$. Thus $m_{T_W}|m_T$.

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    $\begingroup$ Yes, this is correct. $\endgroup$ Commented Jan 17, 2023 at 19:51

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As noted in the comments, your proof is correct. Implicit in your proof is the more general observation that if $W$ is invariant under $T$, then $W$ is also invariant under $f(T)$ for any polynomial $f(x)$ over the base field, and $$f(T)_W=f(T_W)$$ Taking $f=m_T$, we have $$m_T(T_W)=m_T(T)_W=0_W$$ so $m_{T_W}$ must divide $m_T$.

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