I need a way for computing the eigenvalues of the block matrix \begin{equation}R=\begin{bmatrix} 0 & B \\ B^T & D \\ \end{bmatrix}\end{equation} where:
- $B$ is a $1 \times n$ matrix.
- $D$ is a an $n \times n$ matrix
- $0$ is the null $1 \times 1$ matrix
By using https://en.wikipedia.org/wiki/Determinant#Block_matrices, it is known that $$char(R) = \det(-s(D-sI)-B^TB).$$
How to find the roots of char(R)? Is there a formula in terms of eigenvalues of $D$ and $B^TB$?
For instance, let B=[b1,b1,b1,b2,b2,b2,b2], where b1=1/$\sqrt{21}$ and b2=1/$\sqrt{14}$ and D be a matrix with rows [[0,r,r,0,0,0,0],[r,0,r,0,0,0,0],[r,r,0,0,0,0,0],[0,0,0,0,s,0,0],[0,0,0,s,0,0,0],[0,0,0,0,0,0,s],[0,0,0,0,0,s,0]], where r=1/3 and s=1/2. Then the following sage code gives the roots as: [1.000, 0.601, 0.500, -0.333, -0.333, -0.435, -0.499, -0.500]. Note that eigenvalues of D: {2/3,-1/3,-1/3,-1/2,-1/2,1/2,1/2} and $B^TB$: {3/7,0,0,0,0,0,0}.
p=2
b1=1/(sqrt(3*(3+2*p))
b2=1/(sqrt(2*(3+2*p))
B=Matrix(Reals(),[b1,b1,b1,b2,b2,b2,b2])
Id=identity_matrix(Reals(),3+2*p)
r=1/3
s=1/2
D=Matrix(Reals(),[[0,r,r,0,0,0,0],[r,0,r,0,0,0,0],[r,r,0,0,0,0,0],[0,0,0,0,s,0,0],[0,0,0,s,0,0,0],[0,0,0,0,0,0,s],[0,0,0,0,0,s,0]])
s = PolynomialRing(CC, 's').gen()
char_R=(-s*(D-s*Id)-(B.transpose())*B).det()
char_R.roots()
Use online calculator: http://aleph.sagemath.org/
In general, the characteristic polynomial of $R$ will have degree $n+1$, and may be irreducible. For a somewhat random example, take $$ B = (0,1,2,0),\ D = \pmatrix{1 & 0 & -1 & -1\cr -1 & -1 & 2 & 1\cr -1 & 1 & 2 & 0\cr 2 & 2 & 0 & 2\cr}$$ where $R$ has characteristic polynomial $$ t^5 - 4 t^4 - 5 t^3 + 19 t^2 - t - 30 $$ This has Galois group $S_5$, according to Maple, and thus is not solvable by radicals. Therefore the eigenvalues can't be written as expressions (involving the arithmetic operations and radicals) in the eigenvalues of $D$ and $B B^T$ (which are expressible in radicals).