Consider the $2k\times 2k$-matrix $$A= \begin{pmatrix} 0& \frac{1}{k}& \dots& \frac{1}{k}& \frac{1}{k}& 0& \dots& 0\\ \frac{1}{k-1}& 0& \dots& \frac{1}{k-1}& 0& 0& \dots& 0\\ \vdots\\ \frac{1}{k-1}& \frac{1}{k-1}& \dots& 0& 0& 0& \dots& 0\\ \frac{1}{k}& 0& \dots& 0& 0& \frac{1}{k}& \dots& \frac{1}{k}\\ 0& 0& \dots& 0& \frac{1}{k-1}& 0& \dots& \frac{1}{k-1}\\ \vdots\\ 0& 0& \dots& 0& \frac{1}{k-1}& \frac{1}{k-1}& \dots& 0 \end{pmatrix}. $$ Here we have zero on the diagonal, the first and $(k+1)$-th row have row elements $\frac{1}{k}$ and all other rows $\frac{1}{k-1}$.
How do I find the diagonal matrix of $A$? I know that one must find the eigenvalues, but how do I do that for this $2k\times 2k$-matrix?
Edit (my attempts):
I know that the determinant of a matrix with values $a_{ij}$ is given by
$$\det(A)=\sum\limits_{j=1}^{2k}a_{ij}A_{ij}$$
where $A_{ij}$ is the determinant of matrix $A$ with the $i$-th row and $j$-th column left out multiplied by $(-1)^{i+j}$.
I see that we have a lot of zeros and thus many terms will drop out, however I don't see how I can do this in a neat, concise manner.
Edit (further explanation on the matrix):
The right upper square matrix and left lower square matrix are fully zero, except for $a_{1,k+1}=a_{k+1,1}=\frac{1}{k}$.
The other two square matrices have zero only on the diagonal, the first and $(k+1)$-th row have row elements $\frac{1}{k}$ and all other rows $\frac{1}{k-1}$.
So each row sums up to $1$.