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Let $V$ be $n$-dimensional vector space over $F$ and $T\in L(V,V)$. If $T$ is diagonalizable and $c_1,…,c_k$ are distinct eigenvalues of $T$, then $m_T=p=(x-c_1)\dotsb (x-c_k)$.

My attempt: We need to show $p$ is monic, $p(T)=0$ and if $f\in F[x]$ such that $f(T)=0$, then $\deg (p)\leq \deg (f)$. Clearly $p$ is monic. By theorem 3 section 6.3, $c_1,…,c_k$ are roots of $m_T$. By this post, $k\leq \deg (m_T)$. It’s easy to check, $\deg (p)=k$. So $\deg (p)\leq \deg (m_T)\leq \deg (f)$. We claim $p(T)=0$. Since $T$ is diagonalizable, $\exists B=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $\alpha_i$ is eigenvector of $T$, $\forall i\in J_n$. Let $\alpha_i\in B$. Then $\exists c_j\in F$ such that $T(\alpha_i)=c_j\cdot \alpha_i$. Since $F[x]$ is commutative linear algebra over $F$, we have $p=(x-c_1)\dotsb (x-c_k)=(x-c_1)\dotsb (x-c_{j-1})(x-c_{j+1})\dotsb (x-c_k)(x-c_{j})$. So $p(T)=(T-c_1I)\dotsb (T-c_jI)$. Since $(T-c_jI)(\alpha_i)=0$, we have $[p(T)](\alpha_i)=0$. Since $\alpha_i$ was arbitrary, $[p(T)](\alpha_i)=0$, $\forall i\in J_n$. Thus $p(T)=0$. By uniqueness, $m_T=p$. Is my proof correct?

user264745
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    The question is somewhat ambiguously stated. It should make explicit the assumption that $c_1,\dots,c_k$ are all the eigenvalues of $T$. – Ted Shifrin Dec 17 '22 at 22:21
  • @TedShifrin yeah I agree you. In Hoffman and Kunze book word distinct is used that’s why I use it here. – user264745 Dec 18 '22 at 06:46

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While I can see no errors in your argument, I would say it achieves this easy result in a rather roundabout way. Note for instance that you introduce an arbitrary annihilating polynomial $f$ but never really use it in the argument; you go straight for the minimal polynomial $m_T$ instead. Which is valid if you know that the minimal polynomial exists: any monic annihilating polynomial whose degree does not exceed that of the minimal polynomial is the minimal polynomial, though there is an smell of circularity when using that in an argument.

I would use the main fact about polynomials in $T$ and eigenvectors instead, namely that $P[T](v)=P[\lambda](v)$ for any polynomial $T$, whenever $T(v)=\lambda v$. For a polynomial $P(T)$ in a diagonalisable $T$ to be $0$, it suffices that it kills all eigenvectors (for it will then be zero on the sum of eigenspaces, which is the whole space), and for this one must have $P[\lambda]=0$ whenever $\lambda$ is an eigenvalue of $T$. Clearly the minimal degree monic polynomial with this property is the product of factors $X-\lambda$ for $\lambda$ running over all distinct eigenvalues.

Marc van Leeuwen
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  • Thank you so much for the answer. Your points are valid. Hoffman’s define minimal polynomial in this way. So we know existence & uniqueness of minimal polynomial for $T$. Minimal polynomial $p$ for $T$ is uniquely determined by following three properties: (1) $p$ is monic, (2) $p(T)=0$, (3) if $f\in F[x]$ such that $f(T)=0$, then $\deg (p)\leq \deg (f)$. Property (2) & (3) is equivalent to $M_T={f\in F[x] \mid f(T)=0}=pF[x]$. Latter definition has smell of circularity, as you suggested. Second paragraph: your proof is really elegant. – user264745 Dec 20 '22 at 18:25