Characteristic polynomial of a $2d \times 2d$ real symmetric matrix. Let $d \in \mathbb{N}_{ \geq 1}$ denote a dimension and let $p \in (0,1)$. Given the $2d \times 2d$ matrix\begin{align} 
A := \begin{pmatrix}
p & \frac{1-p}{2d-1} & \dots & \dots & \frac{1-p}{2d-1}\\
\frac{1-p}{2d-1} & \ddots & \ddots &  & \vdots \\ 
\vdots & \ddots&\ddots & \ddots & \vdots \\
 \vdots &  & \ddots  & \ddots & \frac{1-p}{2d-1} \\
 \frac{1-p}{2d-1} & \dots & \dots&  \frac{1-p}{2d-1} & p
\end{pmatrix}.
\end{align} 
The above matrix shows up, for example, when studying certain step reinforced random walks. In words, the diagonal entries are saturated by values of $p$, whereas the lower and upper halves of the matrix $A$ all have the same entries $$ \frac{1-p}{2d-1}, $$
in particular these entries depend on the dimension $d$.
One easily verifies that 


*

*For $d=1$, where $A$ reduces to a $2 \times 2$ matrix, the eigenvalues are $ \lambda_1 =1$ and $\lambda_2= 2p-1$.

*For $d=2$, where $A$ reduces to a $4 \times 4$ matrix,  the eigenvalues are $\lambda_1=1$ and $\lambda_2=\lambda_3=\lambda_4=(4p-1)/3$.



Claim: For a general dimension $d$, the eigenvalues of $A$ are given by $\lambda_1=1$, $\lambda_2=\lambda_3= \dots = \lambda_{2d} = \frac{2dp-1}{2d-1}$.


Question: Is there an effective method to compute the characteristic polynomial of a "highly symmetric matrix" as given by $A$, above for a general dimension $d$?
Remarks: A proof by induction seems impossible, as when we increase the dimension, the entries of the Matrix change. On the other hand, computation by the means of Laplace expansion seem complex. 
 A: Proof of the claim: The matrix has the form $\frac{1-p}{2d-1}E + \frac{2dp-1}{2d-1}I_{2d}$, where $E$ is the matrix with all ones. Moreover, $E = 2dP$, where $P$ is the orthogonal projection onto $\operatorname{span}\{(1,1,\ldots,1)^T\}$. Thus
$$
A = \frac{2(1-p)d}{2d-1}P + \frac{2dp-1}{2d-1}I_{2d}
$$
has the set of eigenvalues
\begin{align}
\sigma(A)
&= \left\{\frac{2(1-p)d}{2d-1}\lambda + \frac{2dp-1}{2d-1} : \lambda\in\sigma(P)\right\}\\
&= \left\{\frac{2dp-1}{2d-1},\frac{2(1-p)d}{2d-1} + \frac{2dp-1}{2d-1}\right\}\\
&= \left\{\frac{2dp-1}{2d-1},1\right\}.
\end{align}
A: Hint This matrix is a rank-$1$ adjustment of a scalar matrix, that is, we can write it in the form
$$A = bI + {\bf u}{\bf v}^\top$$ for some column vectors ${\bf u}, {\bf v}$.
In our case, $$b = p - \frac{1 - p}{2 d - 1},$$ and we can, for example, take $${\bf u} := \frac{1 - p}{2 d - 1}\pmatrix{1\\\vdots\\1} , \qquad {\bf v} := \pmatrix{1\\\vdots\\1} .$$
The characteristic polynomial of $A$ is
$$c_A(t) := \det\left[t I - (bI + {\bf u}{\bf v}^\top)\right] = \det\left[(t - b) I - {\bf u} {\bf v}^\top\right] .$$
So, (for generic values of $t$) $c_A$ is the determinant of a rank-one update of an invertible matrix whose determinant we know, which is precisely what the Matrix Determinant Lemma computes:
$$\det (P + {\bf u} {\bf v}^\top) = \left(1 + {\bf v}^\top P^{-1} {\bf u}\right) \det P .$$
But $c_A$ is a polynomial in $t$ and hence continuous, so the polynomial expression given by applying this identity to our above expression for $c_A$ is valid for all $t$.
