Existence of Minimal Polynomial for Square Matrix over Field

Theorem
Let $K$ be a field.

Let $n$ be a natural number.

Let $K^{n \times n}$ be the set of $n \times n$ matrices over $K$.

Let $A \in K^{n \times n}$.

Then the minimal polynomial of $A$ exists and has degree at most $n^2$.

Proof
By Matrices over Field form Vector Space:


 * $K^{n \times n}$ forms a vector space under usual matrix addition and scalar multiplication.

By Dimension of Vector Space of Matrices:


 * $K^{n \times n}$ has dimension $n^2$.

Consider the collection of vectors:


 * $I, A, A^2, \ldots, A^{n^2}$

Since this is a collection of $n^2 + 1$ vectors, and $K^{n \times n}$ has dimension $n^2$, we have by Size of Linearly Independent Subset is at Most Size of Finite Generator:


 * $I, A, A^2, \ldots, A^{n^2}$ are linearly dependent.

That is, there exists $\alpha_0, \alpha_1, \ldots, \alpha_{n^2} \in K$ not all zero such that:


 * $\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i A^i = 0$

That is, the polynomial:


 * $\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i X^i \in K \sqbrk X$

has $\map P A = 0$, and degree at most $n^2$.

Let:


 * $\displaystyle X = \set {P \in K \sqbrk x \setminus \set 0 \mid \map P A = 0}$

$X$ is certainly non-empty since we have found such an element in the computation above.

Now consider the set:


 * $\displaystyle D = \set {\deg P \mid P \in X}$

Since $D$ is a subset of the natural numbers, it contains a least element $N$ by the Well-Ordering Principle.

Since the polynomial we constructed has degree at most $n^2$, we have $N \le n^2$.

Let $Q \in X$ be of degree $N$.

Let $a_N$ be the coefficient of the $X^N$ term in $Q$.

Then $\mu = \dfrac 1 {a_N} Q$ is a monic polynomial of minimum degree with $\map \mu A = 0$.

So $\mu$ is a minimal polynomial for $A$.