How does the idea of a "minimal polynomial" for a matrix (i.e. for a matrix $A$, the polynomial, $\mu (x)$, of least degree, such that $\mu (A) =0$) relate the the "minimal polynomial" for some element, $\alpha $ over a field (i.e. the minimal polynomial for $\sqrt {2}$ over $\mathbb{Q} $ is $x^2-2$)? In the previous example, I'm wondering if there some matrix with entries in $\mathbb{Q} $ which has minimal polynomial $x^2-2$ that somehow relates to $\sqrt {2}$? 


Thanks!



How do minimal polynomials relate?
The general notion of minimal polynomial, as Roy Smith indicated in the comments, works whenever you have an algebra $A$ over a field $K$ and an element $a\in A.$ Namely, consider the unique algebra homomorphism $\pi:k[t]\to A$ that sends $a$ to $t.$\* Since $k[t]$ is a principal ideal domain, its kernel $\ker\pi$ is $(P),$ where $P$ may be chosen to be a monic polynomial of degree $d\geq 0$, or 0. In the former case, $f$ is the minimal polynomial of $a$ and in the latter case, $a$ generates a polynomial algebra inside $A.$ The canonical object which perhaps deserves the name "minimal polynomial" is the ideal $(P)$. When $A$ is a general commutative ring, same reasoning leads to the ideal $\ker\pi$, but there is no reason to expect that it is principal.




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(\*) The idea of "prolonging" the element $a$ to the homomorphism $\pi$ is an algebraic version of the "functional calculus". Thus expressions $q(a)$ for various polynomials $q\in K[t]$ make sense: they are simply $\pi(q).$ However, the functional calculus used in analysis and operator algebra theory involves much larger functional spaces than just polynomials.




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Another useful interpretation of the minimal polynomial is through the poles of the formal resolvent of $a.$ For any commutative ring $A$ (not necessarily an algebra over a field), consider 


$$
(\lambda-a)^{-1}=\sum\_{n=0}^{\infty}\lambda^{-n-1}a^n \in A((\lambda^{-1})).
$$ 


*A priori*, this is just a Laurent formal power series in $\lambda^{-1}$; however, 



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> When $a$ has a minimal polynomial $P(t)$, it is in fact a rational function of $\lambda$ with the denominator (in lowest terms) $P(\lambda),$ and conversely. 
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Indeed,



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> > $P(a)(\lambda-a)^{-1}=P(\lambda)(\lambda-a)^{-1}$ modulo polynomials in $\lambda.{\ } \square$
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See my answer
[How would you solve this tantalizing Halmos problem?](https://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem/31790#31790)
for related ideas.


