General eigenvalues of a matrix for $\bf B_{2n}=\bf 1_{2n}-I_n\otimes \sigma_x$ Let $\bf I_{2n}$ be the unit matrix, $\bf 1_{2n}$ be a $2n\times 2n$ matrix full of $1$ and $\sigma_x=\pmatrix{0&1\\1&0}$. Is it possible to calculate the eigenvalues of a matrix 
$$
\bf B_{2n}=\bf 1_{2n}-I_n\otimes \sigma_x
$$
e.g. $n=2$:
$$
{\bf B_{2\cdot 2}}={\bf 1_4}- {\bf I_2}\otimes \sigma_x=\pmatrix{
1&0&1&1\\
0&1&1&1\\
1&1&1&0\\
1&1&0&1
}
$$
Looks like the only eigenvalues (with multiplicities) are $(-1)^{n-1},(+1)^n$ and $(2n-1)^1$, so characteristic polynomial should $(x-2n+1)(x+1)^{n-1}(x-1)^n$. How to prove that?
 A: Note that the columns of $B_{2n}-I_{2n}$ occur in pairs, i.e. columns $1$ and $2$ are the same, columns $3$ and $4$ are the same, etc.
Hence the rank of $B_{2n}-I_{2n}$ is at most $n$, hence the kernel is at least $n$-dimensional. So the eigenvalue $1$ appears with multiplicity at least $n$. We also get $n$ linear independent eigenvectors:
$$\begin{pmatrix}1\\-1\\0 \\ \vdots\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\\-1\\0 \\ \vdots\\0\end{pmatrix}, \dotsc, \begin{pmatrix}0 \\ \vdots\\0\\1\\-1\end{pmatrix}$$
Now look at $B_{2n}$ again. The column sum of all columns is $2n-1$, hence $2n-1$ is an eigenvalue and the corresponding eigenvector is the unit vector $(1,1, \dotsc, 1)^T$.
We are left to find $n-1$ linear independent eigenvectors with respect to the eigenvalue $-1$, hence we consider the matrix
$$B_{2n}+I_{2n} = \begin{pmatrix}
2I_2 & 1_2 & \dotsc & 1_2 \\
1_2 & 2I_2 & \dotsc & 1_2 \\
\vdots & 1_2 & \ddots  \vdots \\
1_2 & \dotsc & 1_2 & 2I_2
\end{pmatrix}$$
Its pretty obvious that the $n-1$ vectors (The $-1$'s move from $2$nd doublespot to $n$-th doublespot)
$$\begin{pmatrix}1\\1\\-1\\-1\\0 \\ \vdots \\ 0\end{pmatrix}, \dotsc, \begin{pmatrix}1\\1 \\0\\ \vdots \\ 0 \\ -1 \\ -1\end{pmatrix}$$
are contained in the kernel, since if we add up two consecutive columns, we get  a vector full of $2$'s. Furthermore these $n-1$ vectors are clearly linear independent: If we look at the $2(j+1)$-th entry, only the $j$-th vector has a non-zero entry.
Summing up, we have found


*

*$n$ linear independent eigenvectors w.r.t to the eigenvalue $1$,

*$n-1$ linear independent eigenvectors w.r.t to the eigenvalue $-1$,

*$1$ linear independent eigenvector w.r.t to the eigenvalue $2n-1$.


$n+(n-1)+1$ is the size of the matrix, hence we have found all eigenvectors and deduce the characteristic polynomial to be indeed $(x-1)^n(x+1)^{n-1}(x-2n+1)$. The minimal polynomial is $(x^2-1)(x-2n+1)$.
A: Wow, what a great hint of @Daniel. Let me elaborate it:
Let $T$ be a matrix, which diagonalizes $1_n$, i.e. $T^{-1}1_nT=diag(n,0, \dotsc, 0)$.
We note that $1_2\sigma = \sigma 1_2$ and both are diagonalizable, hence they are simultaneously diagonalizable, say by $S$, i.e. we have $S^{-1}1_2S=\begin{pmatrix}2&0\\0&0\end{pmatrix}$ and $S^{-1}\sigma S=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ (to be precise, $S=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ does it).
Now we compute:
$$S^{-1}T^{-1}(1_n \otimes 1_2 - I_n \otimes \sigma)TS = T^{-1}1_nT \otimes S^{-1}1_2S - I_n \otimes S^{-1}\sigma S$$
This is equal to
$$diag(n,0, \dotsc, 0) \otimes \begin{pmatrix}2&0\\0&0\end{pmatrix} - I_n \otimes \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
, which is equal to
$$diag(2n,0, \dotsc, 0)-diag(1,-1,1,-1, \dotsc, 1,-1) = diag(2n-1,1,-1,1, \dotsc, -1,1)$$ and we can see the eigenvalues immediately.
