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  1. Consider $A=\begin{bmatrix} 0 & -2 & 1 \\ 1 & 2 & -1 \\ 3 & -1 & -3 \end{bmatrix}$ . Is $A$ diagonalizable over $\mathbb{C}$?

  2. Does there exist a $3\times 3$ matrix with rational coefficients with no eigenvectors over $\mathbb{Q}$ which is not diagonalizable over $\mathbb{C}$? Find an example of such a matrix, or prove non exists.

For 1), the characteristic polynomial is $\lambda^3+\lambda^2-8\lambda-7$. If I can factor this, and if the eigenvalues are distinct, then $A$ is diagonalizable. However, I don't think I can factor this on my own. I wonder if there is any other easier ways to determine whether this is diagonalizable.

For 2), I don't think such matrix exist, because if it's over $\mathbb{C}$, we can always make the characteristic polynomial split. Is this correct?

Thanks in advance for your help!

RobPratt
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Toasted_Brain
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    Note that you don't need to factor a polynomial $p$ to determine the multiplicity of its roots. Remember that if $r$ is a root of $p$ with multiplicity larger than $1$, then it is also a root of the derivative $p'$. So, you can compute $\gcd(p,p')$ and see if this is $1$. Remember also that you can compute $\gcd$ using Euclid's algorithm. – plop Aug 19 '22 at 00:23
  • For part (2) one would need to have $p$ with a root $r$ of multiplicity $2$ or $3$. Now, the polynomial $\gcd(p,p')$ has rational coefficients (because it came from doing Euclid on polynomials with rational coefficients), has $r$ as root, and has degree $1$ or $2$. So, yes, the root would need to be rational, which is a contradiction. – plop Aug 19 '22 at 00:28

1 Answers1

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  1. If $p=\chi_A$ is reducible over $\mathbb{Q}$, then it has a rational root (because of degree $3<2+2$). By Gauß lemma, it will then have an integral root. Since $p$ is monic, this integral root $\xi$ must devide the constant coefficient $7$, which leaves $\xi\in\{\pm 1, \pm 7\}$ as candidates. None works out. Thus $p$ is irreducible over $\mathbb{Q}$ and has $3$ distinct roots. Thus $A$ is diagonizable.

  2. See the comments.

Thomas Preu
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  • Who says we are looking for rational roots? – Salcio Aug 19 '22 at 01:01
  • Because irreducible polynomials defined over a characteristic 0 field like $\mathbb{Q} are separabel, we would get distinct roots anyway. To get a multiple roots, we need reducibility over $\mathbb{Q}$. – Thomas Preu Aug 19 '22 at 01:09
  • Thomas, the problem above is about polynomial having 3 distinct roots. Not about having rational roots. – Salcio Aug 19 '22 at 11:46
  • @Salcio I agree that the ultimate goal is to show that the roots are distinct. Know that a rational polynomial of degree 3 without rational roots implies that its roots are distinct. Thus showing that it has no rational roots will do the job. – Thomas Preu Aug 19 '22 at 12:50