# Exercise 2, Section 6.4 of Hoffman’s Linear Algebra

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 $$m_{T_W}|m_T$$.

• Yes, this is correct. Commented Jan 17, 2023 at 19:51

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 $$m_T$$.