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Both matrices and Algebraic integers have minimal polynomials. But I struggle to get how these two types of minimal polynomials are related. Well, they are two different rings, but I think we can find some connection between them

For example, if an algebraic number $\alpha$ can be expressed as a combination of the linear independent basis $1,\sqrt{3}, \sqrt{5},\sqrt{15}$, then $\alpha^n$ can be expressed as a combination of this basis as well. The equation $\sum_{i=0}^4 c_i\alpha^i=0$ is essentially four linear equations with four unknowns $c_i$. So, in general, the values of $c_i$ are uniquely determined, and we therefore find the minimal polynomial.

The process above is essentially linear algebra. However, I find that the power $\alpha^n$ cannot be easily represented by the power of a matrix. Is there an easy way to embed the algebraic integers into a set of matrices so that the minimal polynomial of algebraic integers can be found easily by finding the minimal polynomial of matrices?

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  • $\begingroup$ If you know the minimal polynomial $p(x)$ then you can construct a companion matrix which has characteristic polynomial equal to $p(x)$, but not quite what you are looking for. On the other hand there are formulas for the coefficients of $p(x)$ in terms of conjugates of $\alpha$, so there is some relationship I guess? $\endgroup$
    – Yong Hao Ng
    Feb 22, 2019 at 12:17

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Let $\alpha$ be an algebraic integer, let $K={\bf Q}(\alpha)$. Then multiplication by $\alpha$ is a $\bf Q$-linear transformation on $K$. Let $B$ be a basis for $K$ as a $\bf Q$-vector space, and let $M$ be the matrix representing multiplication by $\alpha$ with respect to $B$. Then the minimal polynomial for $\alpha$ over $\bf Q$ is the minimal polynomial for $M$.

Let the degree of $\alpha$ over $\bf Q$ be $n$. Then you can choose the basis $\{\,1,\alpha,\alpha^2,\dots,\alpha^{n-1}\,\}$, and the matrix $M$ that you get has a particularly simple form; it's called the companion matrix for $\alpha$. But it requires that you know how to express $\alpha^n$ in terms of the basis, which is the same as knowing the minimal polynomial of $\alpha$.

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  • $\begingroup$ So what you mean is that matrices cannot help us find minimal polynomial in general. (although it is related) $\endgroup$
    – Ma Joad
    Feb 22, 2019 at 13:04
  • $\begingroup$ I think if you have enough information to find a matrix corresponding to the algebraic integer, then you have enough information to find the minimal polynomial without using the matrix. $\endgroup$
    – Gerry Myerson
    Feb 22, 2019 at 20:47
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    $\begingroup$ In the general case (if $K\neq Q(a)$), the characterisic polynomial of $a$ is a power of its minimal polynomial. See e.g. math.stackexchange.com/a/2710883/300700 $\endgroup$
    – nguyen quang do
    Feb 23, 2019 at 6:59
  • $\begingroup$ By cayley Hamilton theorem, a matrix would give you the characteristic, if it's a field, the minimal polynomial divides this polynomial. $\endgroup$
    – marshal craft
    Jan 14, 2021 at 11:55
  • $\begingroup$ Also I have basically said all these aspects in a less then finished manor, prior to anybody, and as is which dent by nguyens comment, it is contained in my amswer which was most earliest posted. Yet I receive no credit and down votes. Really is that how math people behave? I could even argue that if you do fall into this behavior, there is irratinalality brewing in you which will detriment your ability to mathematically reason. You will never be capable of understanding math, merely mimicking to perfection. $\endgroup$
    – marshal craft
    Jan 14, 2021 at 12:00

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