Linear operator $T$ on $F^2$ defined by $[T]_B=\begin{bmatrix}0&0\\ 1&0 \\ \end{bmatrix}$ has cyclic vector for $T$

Question

In Linear algebra by Hoffman and Kunze,it is given that "an example of an operator which has a cyclic vector is the linear operator $T$ on $F^2$ which is represented in the standard ordered basis by the matrix $\bigl[ \begin{smallmatrix} 0 & 0 \\ 1 & 0 \end{smallmatrix} \bigr]$"

My question is HOW?

My argument:Since the characteristic polynomial of the matrix is $x^2$,so its minimal polynomial is $x$.So,the characteristic polynomial and the minimal polynomial are different.By making use of Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....

Hence,$T$ cannot have the cyclic vector.

Please help!!

P.S.

Please explain stuff in yellow highlighted text!!

@Itay4:beacause it is monic and irreducible and has the same root as that of $x^2$. — Styles, Aug 16 '17 at 12:14
If the minimal polynomial is $x$, then $T=0$, contradiction. — Fred, Aug 16 '17 at 12:16

score 0 · Accepted Answer · answered Aug 16 '17 at 12:19

0

Let $v=(1,0)^T$. Then $Tv=(0,1)^T$. Thus $\{v,Tv\}$ is a basis of $F^2$.

Conclusion: $v$ is a cyclic vector.

answered Aug 16 '17 at 12:19

Fred

77,394

1

This is not an answer to the question. The OP wanted to know what was the error in his argument. – José Carlos Santos Aug 16 '17 at 12:27
And that is reason for the downvote ???? In a comment above I told the OP what was the error, and then , in my answer, I showed the existence of a cyclic vector. Please read the question " My question is HOW?" of the OP and don't be so petty. Thanks ! – Fred Aug 16 '17 at 12:38
I did not downvote your answer. – José Carlos Santos Aug 16 '17 at 13:14
I beg your pardon. Who else downvoted my answer ? – Fred Aug 16 '17 at 13:19

user264745 · Answer 2 · 2023-02-17T21:24:11.250

Characteristic polynomial of matrix $\begin{bmatrix}0&0\\ 1&0\\ \end{bmatrix}$ is $x^2$. So it’s minimal polynomial is $x$ or $x^2$. Certainly $x$ is not minimal polynomial, because $x$ don’t annihilate that matrix. Thus $x^2$ is minimal polynomial. The theorem you stated is correct but we can’t use result of that theorem, since it’s proof comes later in the book.

Author explicitly showed $e_1$ is cyclic vector of $T$, i.e. $F^2=Z(e_1;T)=\{g(T)(e_1)\mid g\in F[x]\}$. Let $\beta =(a,b)\in F^2$. Note $T(e_2)=(0,0)$. So $\beta =a\cdot e_1+b\cdot e_2=a\cdot e_1+b\cdot T(e_1)$. Take $g=a+bx$. Clearly $g(T)(e_1)=\beta \in Z(e_1;T)$. Thus $F^2\subseteq Z(e_1;T)$. Trivially $F^2\supseteq Z(e_1;T)$. Hence $F^2=Z(e_1;T)$.

Since $T(e_2)=(0,0)=0\cdot e_2$, we have $e_2$ is eigenvector of $T$ corresponding to eigenvalue $0$. By example 1 section 7.1, $\dim Z(e_2;T)=1$. Thus $Z(e_2;T)\neq F^2$. Hence $e_2$ is not cyclic vector of $T$.

Edit: Well, I recently proved the theorem OP used. It’s proof is not trivial.

Linear operator $T$ on $F^2$ defined by $[T]_B=\begin{bmatrix}0&0\\ 1&0 \\ \end{bmatrix}$ has cyclic vector for $T$

2 Answers2