Prove square matrix over $\mathbb{Z}$ can't have $\frac{1}{4}(-3+ i \sqrt5)$ as an eigenvalue.

My proof:

  • If matrix has eigenvalue z=$\frac{1}{4}(-3+ i \sqrt5)$, then it must has eigenvalue $\frac{1}{4}(-3- i \sqrt5)$ (conjugate)
  • So, the characteristic polynom has form P=$(x-z)(x - \overline z) Q$ , where Q - some polynom over $\mathbb{R}$
  • $P= a_0x^n + a_1x^{n-1} + a_2x^{n-2} \dots + a_n$, where $a_0 \dots a_n \in \mathbb{Z}$ (because matrix over $\mathbb{Z}$)
  • $Q= q_0x^{n-2} + q_1x^{n-3} + q_2x^{n-4} \dots + q_{n-2}$, where $q_0 \dots q_n \in \mathbb{R}$
  • $(x-z)(x - \overline z) = (x-\frac{1}{4}(-3+ i \sqrt5))* (x-\frac{1}{4}(-3- i \sqrt5)) = x^2 + \frac{3}{2}x + \frac{7}{8}$
  • $q_0=1$, otherwise $P$ is not a characteristic polynom($a_0 \neq 1$)

  • But if $q_0=1$, then $a_1= \frac{3}{2}$ , contradiction

Am i right?

  • 1
    $\begingroup$ $a_1 = {3 \over 2} + q_1$ $\endgroup$ – Abstraction May 26 '16 at 12:23
  • $\begingroup$ Yes, you are right. Does you have any idea to prove the theorem? $\endgroup$ – Evgeny Semyonov May 26 '16 at 12:30

Perhaps there's an easier, direct way, but isn't this a matter of using Gauss' Lemma?

We use two facts.

  1. the product of two primitive polynomials is primitive
  2. if $f$ is a non-zero rational polynomial, then $f = \gamma f_{1}$, where $\gamma$ is a rational number, and $f_{1}$ is a primitive polynomial with integer coefficients. The decomposition is unique up to signs

So $$ f = x^2 + \frac{3}{2}x + \frac{7}{8} = \frac{1}{8} (8 x^{2} + 12 x + 7), $$ where $f_{1} = 8 x^{2} + 12 x + 7$ is primitive, and $\gamma= 1/8$.

Now note that $Q$ has rational coefficients. (You obtain it by dividing $P$ by $x^2 + \frac{3}{2}x + \frac{7}{8}$.)

Write $Q = \delta Q_{1}$ as above, then $$ P = \gamma \delta f_{1} Q_{1}, $$ where $f_{1} Q_{1}$ has integer coefficients and it is primitive, and thus by uniqueness $\gamma \delta = \pm 1$, so that $\delta = \pm 8$, so that $8$ divides $q_{0}$.

  • $\begingroup$ Thanks! I did't know Gauss' Lemma. One question: why Q must has rational coefficients? $\endgroup$ – Evgeny Semyonov May 26 '16 at 13:56
  • $\begingroup$ You're welcome. As to $Q$, it is probably simpler than this, however: divide $P$ by $f = x^2 + \frac{3}{2}x + \frac{7}{8}$ to get $P = f Q + r$, with $Q, r$ with rational coefficients, and $r$ either 0, or of degree at most $1$. But clearly $r$ has two distinct roots $z, \bar{z}$, so it cannot have degree $0$ or $1$, so $r = 0$. $\endgroup$ – Andreas Caranti May 26 '16 at 14:34

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