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.


So someone already commented my post with an answer but deleted it straight after so I decided to write it here.

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 $b$.


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