Degree of minimal polynomial of a square matrix no larger than dimension

Question

Basic fact: for an $n$-dim space $V$ and nonzero $A\in L(V)$, let $p_A$ be its minimal polynomial, then $\deg(p_A)\leq n$. However, I’ve only seen a proof based on induction, and I can’t seem to modify it to avoid induction. So I wonder if there is a non-induction proof. Thanks in advance.

Answer 1

We have to go through a bit of definitions and theorems. Let us denote with $\mathbb K^{n,n}$ the space of $n\times n$ matrices with coefficients in the field $\mathbb K$. Given $A\in\mathbb K^{n,n}$ we define the ideal of the matrix $A$ as $$I_A:=\{f\in \mathbb K[t]\:|\:f(A)=0\}$$ where $\mathbb K[t]$ is the ring of polynomials with coefficients in $\mathbb K$ and the evaluation of a polynomial $f$ in a matrix $A$ is quite obvious: if $f(t)=b_nt^n+b_{n-1}t^{n-1}+\ldots +b_1t+b_0$ then $$f(A)=b_nA^n+b_{n-1}A^{n-1}+\ldots +b_1A+b_0I\in \mathbb K^{n,n}$$ $I_A$ satisfies some obvious properties:

1. $0\in I_A$
2. $I_A\neq \{0\}$; infact, if we consider the $n^2+1$ matrices in $\mathbb K^{n,n}$: $$I,A,A^2,A^3,\ldots, A^{n^2}$$ then they are linearly dependent, since $\text{dim}\mathbb K^{n,n}=n^2$. So there must exist $a_0,\ldots,a_{n^2}\in\mathbb K$ not all equal to zero such that $$a_0I+a_1A+a_2A^2+\ldots+a_{n^2}A^{n_2}=0$$ Then $p(t)=a_{n^2}t^{n^2}+\ldots+a_2t^2+a_1t+a_0$ is a non-zero polynomial that is in $p(t)$
3. $I_A$ is obviously a subgroup of $(\mathbb K^{n,n},+)$, and then if $f\in I_A$ and $h\in \mathbb K[t]$: $$(fh)(A)=f(A)h(A)=0$$ since $f\in I_A$. So $I_A$ is an ideal of $\mathbb K[t]$.

It's a well-known fact from abstract algebra that, being $\mathbb K$ a field, the ring of polynomials $\mathbb K[t]$ over $\mathbb K$ is a principal ideal, i.e. if $I\subseteq \mathbb K[t]$ is an ideal then it exists $p$ such that $I=(p)$. The generator $p$ is unique up to scalar multiplication (with the scalar being different from zero). So, if we want $p$ to be monic then $p$ must be unique. In the proof of this fact (let me know if you want me to add the proof or if it's clear enough) it's obvious that in $I$ there can't be a polynomial with a degree lower that the one of $p$.

At this point, given $A\in\mathbb K^{n,n}$ and its associated ideal $I_A$, the minimal polynomial $m_A(t)$ of $A$ is defined to be the monic generator of $I_A$, i.e. the monic, non-vanishing, polynomial of minimum degree from the polynomials of $I_A$.

The Cayley-Hamilton theorem (it can be proven without induction) tells us that the characteristic polynomial $C_A(t)$ of $A$ belongs to $I_A$, that is to say $C_A(A)=0$. But, since $I_A$ is generated by the minimal polynomial, this implies that $m_A(t)$ divides $C_A(t)$. A consequence is that $$\text{deg}\:m_A(t)\leq \text{deg}\:C_A(t)=n$$

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EDIT: To see that $C_A(t)$ has degree $n$ just look at this answer.