0
$\begingroup$

I want to calculate the eigenvalues of some diagonalisable linear operator $L$ in some basis $B$ with the following transformation matrix.

$$ [L]_B = \begin{bmatrix} 3+\frac{1}{2}i & -\frac{1}{2}-2i &0 \\ \frac{1}{2}+2i &3+\frac{1}{2}i & 0 \\ 0 & 0 & 1-i \end{bmatrix} $$ I know that the trace of any diagonalisable operator $L$ is equal to the sum of its eigenvalues independent of what basis we choose. So in this case, the sum of eigenvalues is 7. However, I want to calculate the individual values of the eigenvalues. When trying to determine the roots of the characteristical polynomial, I find a very difficult polyniomal that I can't solve by hand. I was thinking that there has to be a smarter way to solve this problem. Could anyone help me?

Thanks for your time,

K. Kamal

$\endgroup$
0
$\begingroup$

Your matrix is block-diagonal, which makes computation easy in this case. In particular, we know that $1-i$ will be the eigenvalue corresponding to eigenvector $(0,0,1)$, and the remaining eigenvalues are simply the eigenvalues of the submatrix $$ \pmatrix{3+ \frac 12 i & - \frac 12 - 2i\\ \frac 12 + 2i & 3 + \frac 12 i} $$ Perhaps you could take it from there.

$\endgroup$
  • $\begingroup$ I think that that is what I was missing. If $1-i$ is an eigenvalue, that makes $1+i$ an eigenvalue as well. But since the sum of the eigenvalues needs to be 7 and we have a $3x3$ matrix, the last eigenvalue needs to be $5$, if I'm correct. $\endgroup$ – K.Kamal Nov 6 '17 at 19:12
  • $\begingroup$ @K.Kamal It is not necessarily the case that $1 + i$ is an eigenvalues. Because this is a matrix with complex (non-real) entries, it is possible to have eigenvalues that do not come in conjugate pairs. $\endgroup$ – Omnomnomnom Nov 6 '17 at 19:14
  • $\begingroup$ @K.Kamal However, it does appear that, in this case, you have arrived at the correct answer $\endgroup$ – Omnomnomnom Nov 6 '17 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.