Calculating the eigenvalues of a diagonalisable linear operator $L$.

Question

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

Answer 1

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.