Minimal polynomial

Question

Let $V$ be the vector space of square matrices of order $n$ over the field $F$. Let $A$ be a fixed square matrix of $n$ and let $T$ be a linear operator on $V$ such that $T(B) = AB$. Show that the minimal polynomial for $T$ is the minimal polynomial for $A$.

Thank you for your time.

by cayley hamilton theorem $p(A)=0$, and $p(T)$ is polynomial in operator and $Tv=kv$, k is an eigen value. I can prove $p(T)=0$. — prasad, Sep 01 '13 at 19:04

score 4 · Accepted Answer · answered Sep 01 '13 at 19:04

4

You need to show that every polynomial that kills $A$ kills $T$. But then to show minimality, you need to show that every polynomial that fails to kill $A$ fails to kill $T$.

$$f(A)=0$$

$$f(T)B = \left(\sum_{k=0}^n c_k T^k \right)B = \sum_{k=0}^n c_k (T^k B).\tag1$$

$$ T^k(B) = T^{k-1}(T(B)) = T^{k-1}(AB) = T^{k-2} (A(AB)) = T^{k-2} (AA(B)) = \cdots. $$ In other words, show by induction on $k$ that $T^k (B) = A^k B$ and then apply $(1)$.

That should suggest how to do the other part.

answered Sep 01 '13 at 19:04

Michael Hardy

1

nice answer. thank you Michael Hardy. – prasad Sep 01 '13 at 19:15
Sir, I have written in detail for characteristic polynomial case, please look at my answer and suggest me changes if necessary. Thank you.. please provide some idea to prove minimal polynomials are same. – Sep 28 '13 at 06:48

score 1 · Answer 2 · answered Dec 04 '14 at 14:40

On checks immediately that $T^k(B)=A^k\cdot B$ for all $k$ and $B$, and so by linearity $P[T](B)=P[A]\cdot B$ for all polynomials$~P$.

Now if $P[A]=0$ then for all $B$ one has $P[T](B)=P[A]\cdot B=0\cdot B=0$ so every polynomial annihilating $A$ annihilates $T$. Conversely if $P[T](B)=0$ for all $B$ then taking $B=I$ gives $0=P[T](I)=P[A]\cdot I=P[A]$, so so every polynomial annihilating $T$ annihilates $A$. Thus their minimal polynomials are the same.

score 1 · Answer 3 · answered Sep 28 '13 at 06:46

This content is actually supposed to be a "Proof verification Question".I thought of posting this as a question but then i realized there is a page having same question. So, I thought to make this as a "partial answer" and ask for help instead of getting a "duplicate question tag"

I have $V$- vector space of all $n\times n$ matrices over a field $F$ and let $A$ be a fixed matrix in $V$.

Define $T : V\rightarrow V$ as $T(B)=AB$

Question is to prove that minimal polynomial of $T$ and $A$ are same.

Suppose $f(x)=\sum_{i=0}^n a_i x^i $ be characteristic polynomial for $T$

i.e., $\sum_{i=0}^n a_i T^i=0$ i.e., $\sum_{i=0}^n a_i T^i(B)=0$ for all $B\in V$

in particular, $\sum_{i=0}^n a_i T^i(I)=0$

Now, $T(I)=A$ and $T^2(I)=T(T(I))=T(A)=AA=A^2$ for similar reasons, $T^n(I)=A^n$

So, we have $\sum_{i=0}^n a_i T^i(I)=0$ implies $\sum_{i=0}^n a_i A^i=0$.

So, $f(x)$ is characteristic polynomial for $A$

Suppose $f(x)=\sum_{i=0}^n a_i x^i $ be characteristic polynomial for $A$

i.e., $\sum_{i=0}^n a_i A^i=0$ but, $T^i(I)=A^i$ So, $\sum_{i=0}^n a_i T^i(I)=0$ i..e, $(\sum_{i=0}^n a_i T^i)(I)=0$

multiplying by arbitrary $B\in V$ we get $(\sum_{i=0}^n a_i T^i)(I)(B)=0.B=0$ i.e., $(\sum_{i=0}^n a_i T^i)(B)=0$ for all $B\in V$ Thus, $f(x)$ is characteristic polynomial for $T$

So, we concluded that characteristic polynomials of $T$ and $A$ are same.

I did not understand how to proceed to show its minimal polynomials are same.

$T$ is a linear operator on $n^2$-dimensional vector space, so characteristic polynomial of $T$ must have degree $n^2$. However $A$ is $n\times n$ matrix, so its characteristic polynomial has degree $n$. Thus, their characteristic polynomial cannot be equal. — Sungjin Kim, Sep 28 '13 at 07:07
@i707107 : nice observation.. i dont know how i missed it.. :( — , Sep 28 '13 at 11:30

score 0 · Answer 4 · answered Sep 23 '15 at 07:44

Let $T$ be a linear operator(or a matrix $A$ over the field $\mathbb{F}$) on a vector space $V(\mathbb{F})$, then the minimal polynomial of $T$ (or of matrix $A$) is the monic generator of the P.I.D. $$\{p(x)\in\mathbb{F}[x]:p(T)=0\}$$ Now just show that $$\{p(x)\in\mathbb{F}[x]:p(T)=0\}=\{p(x)\in\mathbb{F}[x]:p(A)=0\}$$ So that they have same monic generators i.e. same minimal polynomials.

Minimal polynomial

4 Answers4