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3 votes

$\operatorname{rank}(A)=r$, then $I,A,\cdots, A^{r+1}$ is linearly dependent

You can do a bit better: without using $I$, those positive powers of $A$ are already linearly dependent. Let $V$ be the image (column space) of $A$, then by definition of the rank one has $\dim(V)=r$, ...
Marc van Leeuwen's user avatar
3 votes
Accepted

find minimal polynomials

Everything happens inside the field $\Bbb{Q}(\alpha)$. It is a 3-dimensional space spanned by the $\Bbb{Q}$-basis $\mathcal{B}=\{1,\alpha,\alpha^2\}$. One way of seeing that is to check that none of ...
3 votes
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Classify, up to similarity, the $3$ by $3$ matrices with coefficients in $\mathbb{Q}$ that satisfy $A^6=I$.

As $\deg m(x) \leq 3$, the following are your candidates for the minimal polynomial: Degree 1: $x-1$, $x + 1$. Degree 2: $x^2 + x + 1$, $x^2 - x + 1$, $x^2 - 1$. Degree 3: $(x-1)(x^2 + x + 1)$, $(x-...
koifish's user avatar
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2 votes

find minimal polynomials

For the second part, in addition to Jyrki's answer, there are the following two amazing alternative methods to find the minimal polynomial: Compute $f(x)=Res(x^3+2x-1, y-x^2-x)$ with respect to the ...
nor's user avatar
  • 111
2 votes
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Degree of field extension $\Bbb Q(\sum\limits_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}):\Bbb Q$

The answer to your question is yes when $n$ is squarefree. Here is a general result. When $L/K$ is Galois with Galois group $G$, $L = K(\alpha)$, and $H$ is a subgroup of $G$, then we can ask ...
KCd's user avatar
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1 vote

Classify, up to similarity, the $3$ by $3$ matrices with coefficients in $\mathbb{Q}$ that satisfy $A^6=I$.

Another perspective on this question, this is asking for the number of isomorphism classes of representations of the cyclic group of order $6$ of dimension $3$. By general theory, all such ...
Chris H's user avatar
  • 6,870
1 vote

Classify, up to similarity, the $3$ by $3$ matrices with coefficients in $\mathbb{Q}$ that satisfy $A^6=I$.

As mentioned in the comment, the minimal polynomial of a $3\times 3$ matrix may have degree less than $3$. And even if the minimal polynomials are the same, the matrices might still not be similar. ...
Just a user's user avatar
  • 16.1k

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