# Calculating eigenvalues of specific matrix knowing the determinant and that the eigenvalues are complex

The question asks to find eigenvalues of matrix $$M= \begin{pmatrix} 2&0&0&-1\\ 0&2&1&1\\ 1&-1&2&0\\ 1&0&0&2 \end{pmatrix}=\begin{pmatrix} A&B\\ C&A \end{pmatrix}$$ We are given that $$\det(M)=25$$ and there are 2 eigenvalues, both complex.

So I know I can do it from the definition, not using what I am given however I know that there is some trick to do that more neatly. What I did is:

I noticed that this is a block matrix of 2x2 matrices $$A$$ is diagonal so it commutes with $$C$$ whence the determinant of $$M-\lambda I$$ is $$\det((A-\lambda I)^2-BC)$$ and using this formula I got eigenvalues to be $$1\pm 2i$$

However I found out numerically (using this website http://www.bluebit.gr/matrix-calculator/calculate.aspx) that the eigenvalues are $$2\pm i$$.

One has to be wrong, right?

Anyway what I am really interested in is how to solve this problem using what I have been given i..e. that $$\det(M)=25$$ and the eigenvalues are complex

You made a mistake in calculating $$\det((A-\lambda I)^2-BC)$$. What you have is $$BC=\begin{pmatrix}-1\\2&-1\end{pmatrix}$$ so $$(A-\lambda I)^2-BC=\begin{pmatrix}(2-\lambda)^2+1\\ -2 & (2-\lambda)^2+1\end{pmatrix}$$ giving eigenvalues $$2\pm i$$ each with algebraic multiplicity 2.
Trace is invariant over basis change and we can put $$M$$ in Jordan Canonical form where on the diagonal we have only eigenvalues we know there will be 2 of them so they must be conjugates and so $$8=Tr(A)=\sum_{i=1,2}\Re(\lambda_i)$$ so if $$\lambda =a\pm bi$$ we have $$4a=8$$ therefore $$a=2$$ now using the fact determinant is invariant and equal to $$25$$ we can find $$b$$.