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Let $V$ be a finite dimension vector space. $T\in \mathscr L(V)$ and $v\in V$.

Let $S_v = \{ P\in \mathbb C [X], P(T)(v)=0\}$

And let $P_v \in S_v$ be the monic polynomial of lowest degree of $S_v$.

Apparently $P_v$ is the "minimal polynomial" relative to $T(v)$. But I can't find the connection between the definition of a minimal polynomial and $P_v$.

$\mathbb C$ sure is a field. But according to the definition of a minimal polynomial, $T(v) \in V$ so $V$ should be an extension field of $\mathbb C$. I don't see how it is the case. $\mathbb C$ is not even included in $V$.

Can anyone tell me where I am getting it wrong?

Thanks.

niobium
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    Remove the v, what do you get? Also $P(T)$ is a linear endomorphism, not a polynomial. You don’t apply $P$ to $T(v)$ but $P(T)$ to $v$ – julio_es_sui_glace Nov 24 '24 at 15:25
  • Are you saying that $P_v$ does not meet the requirements for being a minimal polynomial? – niobium Nov 24 '24 at 15:40
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    Minimal polynomial for an element in a field extension has a different meaning than minimal polynomial in the context of vector space and linear transformations over a field. Though the definitions are almost identical, and the terminology is the same, the context is different. – Anuradha N. Nov 24 '24 at 15:40
  • @AnuradhaN. Yes but I think the linear algebra definition here is not suited since no endomorphism (nor matrix) is being canceled here – niobium Nov 24 '24 at 15:43
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    You can take the matrix associated to the linear transformation and think of it as a polynomial equation in the matrix. (Also, $P(T)$ is a linear endomorphism from the space ${\cal{L}}(V)$ to itself). – Anuradha N. Nov 24 '24 at 15:47
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    The constant term in the polynomial corresponds to the constant multiplied by the identity matrix (or identity transformation). – Anuradha N. Nov 24 '24 at 15:50
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    What do you mean by requirements? Minimal polynomials are easily defined over any algebra. For field extension it is just a particular case of $k-algebra$ – julio_es_sui_glace Nov 24 '24 at 16:16
  • @julio_es_sui_glace I get your first comment, but the problem is the same for $v$: the vector $v$ belongs to $V$ and how can we see $V$ as a field extension of $\mathbb C$ ($\mathbb C$ being the field over which the polynomial takes its coefficients)? To reply to your question, I am just asking if you say that $P_v$ is not a minimal polynomial – niobium Nov 24 '24 at 16:34
  • Why do you want to see $V$ as a field extension? The algebra you are working with is $\mathbb{C} [T] \subset \mathcal L (\mathbb C )$ – julio_es_sui_glace Nov 24 '24 at 16:51
  • @julio_es_sui_glace Well I was assuming it was a minimal polynomial in the field theory "sense". I based myself on this question. Could it be that $P_v$ is a minimal polynomial in the linear algebra sense? But as I stated to Anuradha N. there isn't any endomorphism (nor matrix) that's being canceled here. What's canceled is its value at $v$. – niobium Nov 24 '24 at 17:02
  • Yes this is why it is « relative to v », you can see this as a minimal polynomial through the prism of $\operatorname{ev}_v$ – julio_es_sui_glace Nov 24 '24 at 17:24
  • Wikipedia says: "More formally, a minimal polynomial is defined relative to a field extension $E/F$ and an element of the extension field $E/F$". If it is relative to $v\in V$ then $V$ should be seen as a field extension? Also, what does $ev_v$ stand for? I am not familiar with algebra over fields (just quickly checked the encyclopedia page). – niobium Nov 24 '24 at 17:34
  • Let us continue this discussion in chat. – julio_es_sui_glace Nov 24 '24 at 17:36

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You appear to be confusing two variations of the notion of minimal polynomial of a vector space endomorphism, both defined in terms of a specific element but of different algebraic structures.

The simplest (to define) notion of minimal polynomial is that of a vector space endomorphism$~\phi$ of some vector space$~V$ over a field$~K$: it is the monic generator of the kernel of the substitution map $X:=\phi$ from $K[X]$ to the $K$-algebra $\mathcal L(V,V)$ of endomorphisms of$~V$. In other word it is the minimal degree monic polynomial $P$ such that $P[\phi]=0\in\mathcal L(V,V)$.

When $v\in V$ is a particular vector, one could instead just require that $P[\phi](v)=0\in V$, which in general may have a lower degree monic solution that the minimal polynomial (it will always divide the minimal polynomial). This is the $P_v$ of your question. It is equal to the minimal polynomial of the restriction of $\phi$ to the smallest $\phi$-stable subspace containing $v$, namely the subspace spanned by the vectors $\phi^k(v)$ for $k\in\Bbb N$.

Finally when $a$ is an element of a field extension $L/K$, the minimal polynomial of $a$ over $K$ is the minimal polynomial of the $K$-vector space endomorphism of$~L$ given by multiplication by$~a$. This is the monic generator of the kernel of the substitution map $X:=a$ from $K[X]$ to the $K$-algebra $L$, in other words the minimal degree monic polynomial $P\in K[X]$ such that $P[a]=0\in L$. There is no direct relation with the polynomials $P_v$.

Marc van Leeuwen
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