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 2×2 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
Best Answer
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.