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I've been trying to solve this question for some time now:

If $A$ is a $3×3$ matrix over $\mathbb{Q}$ that has exactly two eigenvalues over $\mathbb{Q}$ is it triangularizable over that field?

Before that I had the same question but asks about diagonalizability instead of triangularizibility and I found a counter example: $$ \begin{matrix} 1 & 2 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{matrix} $$

$\lambda_{1} = 1$ and $\lambda_{2} = 2$ are the eigenvalues of this matrix, it is not diagonalizable because it only has two linearly independent eigenvectors, I think that the answer to this question is also a counter example, because a matrix is triangularizable if and only if all of its eigenvalues exist over its field, in this case $\mathbb{Q}$, but I might be wrong because there might be multiplicities of the eigenvalues

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    $\begingroup$ If a $3 \times 3$ matrix over $\Bbb Q$ has $2$ eigenvalues in $\Bbb Q$, then all of its eigenvalues must be in $\Bbb Q$! (The characteristic polynomial will factorise completely). $\endgroup$ Apr 3, 2023 at 14:46
  • $\begingroup$ @IzaakvanDongen could you please explain more? $\endgroup$
    – T-Caster
    Apr 3, 2023 at 15:04
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    $\begingroup$ The sum of the roots of a monic polynomial is one of the coefficients of that polynomial (up to a negative sign). What can you say if the coefficients of a cubic polynomial are rational and two of the roots are rational? $\endgroup$ Apr 3, 2023 at 15:57
  • $\begingroup$ What Greg writes is a good way to see it. In some sense what I had in mind is even less "clever" - we know that if a polynomial $P(X)$ over a field $K$ has a root $\alpha$ in $K$, then we can write $P(X) = (X - \alpha) \cdot Q(X)$, where $Q(X)$ is also a polynomial over $K$ ("the factor theorem"). What I wrote already follows from this fact, since you can then write the characteristic polynomial as $(X - \lambda_1)(X - \lambda_2) \cdot Q(X)$, where $Q(X)$ has degree $1$. This way of thinking may be helpful for some other, similar problems. $\endgroup$ Apr 12, 2023 at 17:49

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I solved this question, here's the answer: It's given that $A \in M_{3}(\mathbb{Q})$ therefore $\Delta _{A}$ (It's characteristic polynomial) is a $3^{rd}$ degree, so it looks like this: $$ax^3 + bx^2 + cx +d$$ and according to the definition of $\Delta _{A}$ "b" is the trace of A, which means $b \in \mathbb{Q}$, let's contradictly assume that A is not triangularizable, which means it must have an eigenvalue $z \notin \mathbb{Q}$, it's also given that A has two eigenvalues $\lambda_{1}, \lambda_{2} \in \mathbb{Q}$, which means we can write: $$\Delta_{A} = (x-\lambda_{1})(x-\lambda_{2})(x-z)$$ and if we simplify this expression we get:$$x^3 - (\lambda_{1} + \lambda_{2} + z)x^2 ...$$ as we can see, according to the calculations $b = - (\lambda_{1} + \lambda_{2} + z)$ and for it to be a rational number $z$ must be rational, which contradicts with our assumption, which means that the only possible way for A to have exactly to eigenvalues in $\mathbb{Q}$ is when the multiplicity of one of them is 2, which means it is triangularizable.

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