$$A = \begin{bmatrix} 0 & i+1 & 0 \\ i & 0 &2 -i \\ 2-i & 0 & i \end{bmatrix}$$

and eigenvalues, eigenvectors are asked.

  1. by the frequently used method ,we get the characteristic equation as $$\lambda^{3} - i (1 + \lambda )\lambda + \lambda - 8 = 0$$ no way to solve this equation by hand. By using some WEB sites 3 different complex roots are obtained, all with "ugly" coefficient like $(-1.0463-1.6363i),\dots$ Then by using these eigenvalues, almost impossible to get the eigenvectors by hand.

  2. Diagonalizing matrix $A$ by row operations and picking up the diagonal entries.

I've done this too. I've obtained "nice" eigenvalues but they do not satisfy the original characteristic equation given above. So there is some inconsistance.

What I've missed here? Any help will be appreciated.

(This was an exam question, so it is supposed to be solved in a limited time by hand)

  • $\begingroup$ With a number of elements being zero, and the matrix being quite small perhaps you can use that the trace is the sum of eigenvalues and the determinant is the product of eigenvalues? $\endgroup$ – Benedict W. J. Irwin Jun 30 '20 at 11:00
  • 2
    $\begingroup$ I don't obtain this characteristic polynomial. $\endgroup$ – Bernard Jun 30 '20 at 11:01
  • $\begingroup$ $\lambda^3-i\lambda^2+(1-i)\lambda+8=0$ is the characteristic equation. Regardless, it has ugly solutions! $\endgroup$ – Sameer Baheti Jun 30 '20 at 12:19
  • $\begingroup$ Therefore there might be a typo in the matrix. $\endgroup$ – Dietrich Burde Jun 30 '20 at 12:33

The characteristic polynomial is given by $$ \chi(t)=t^3 - it^2 + ( 1 - i)t - 8 $$ and the roots $z_1,z_2,z_3$ satisfy $$ z_1+z_2+z_3=i,\; z_1z_2z_3=8. $$ So we have $z_3=i-z_1-z_2$ and $z_1z_2(i-z_1-z_2)=8$. This gives a quadratic equation.


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