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

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?

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.
• 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. – K.Kamal Nov 6 '17 at 19:12
• @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. – Omnomnomnom Nov 6 '17 at 19:14