0
$\begingroup$

Given matrix $$M=\begin{bmatrix}0&1&2&0\\ 1&0&1&0\\2&1&0&2\\0&0&2&0\end{bmatrix}$$ Then,

  1. $M$ has no real eigenvalues.
  2. all real eigenvalues of $M$ are positive.
  3. all real eigenvalues of $M$ are negative.
  4. $M$ has both positive and negative real eigenvalues.

So I calculated the characteristic polynomial of $M$ and found it to be $x^4-10x^2-4x+4=0$. I tried using rational root theorem to determine its roots but could not find any. So this means that ooption 1 is correct. But on the other hand every real symmetric matrix has real eigenvalues. So isn't it a absurd conclusion.

  • 0
    Computing the values of the characteristic polynomial at $x=-1$, $x=0$, $x=+1$, shows that 4. and only 4. holds. – 2017-01-04
  • 0
    $p(0)=4, p(1)=-9,p(-1)=-1$ So there are positive and negative real eigenvalues. – 2017-01-04

2 Answers 2

8

First note that as $M$ is symmetric so all our eigenvalues are real. Now we know that $tr(M)=\sum_i \lambda_i $ and $det(M)=\prod_i \lambda_i$. And we have $det(M)=4$ which implies that none of the eigenvalue is zero and combining this with $tr(M)=0$ tells us that option 4 is correct.

  • 2
    Since $M$ is symmetric (all eigenvalues are real), non-zero (not all eigenvalues are zero), and traceless (but the sum of the eigenvalues is zero), there is no need to also compute the determinant to reach the same conclusion. – 2017-01-04
  • 0
    Actually this answer is wrong since it does not use the symmetry of M which is crucial to conclude with this approach. – 2017-01-05
2

The rational root theorem giving no roots for $\chi_M$ does only say that $M$ does not have rational eigenvalues. To determine the signs of the real(! - $M$ is symmetric, hence all its eigenvalues are real) eigenvalues, compute $\chi_M(0)$, $\chi_M(-1)$ and $\chi_M(1)$. We have $$ \chi_M(\pm 1) < 0, \chi_M(0) > 0 $$ So $\chi_M$ has roots in $(0,1)$ and in $(-1,0)$. So 4. is true.