You can just write down the eigenvectors.
For instance, the vector $(1,1,\ldots, 1)$ is transformed into $1 + cp$ times itself, so $1 + cp$ is an eigenvalue.
Similarly, the vector $v_j = e_1 - e_j$ ($j = 2, \ldots, p$)$ is an eigenvector of eigenvalue 1.
So the eigenvalues are:
- $\lambda = 1$, with multiplicity $p-1$
- $\lambda = 1 + cp$, with multiplicity 1.
WAIT! You changed the question while I was writing the answer, from
$$
X_{p\times p} = I_{p\times p} + c\mathbf{1}_{p\times p}
$$
to
$$
X_{p\times p} = I_{p\times p} + c\mathbf{1}_{p\times p} - c \text{diag}(\mathbf{1})
$$
So my answer is to a slightly different question; I leave it to you to adjust things to make it work for your modified question.
Answer to modified question
Your matrix can be rewritten as
$$
M = (1 - c) I + c 1_{p \times p}
$$
I'm going to find the eigenvalues by inspection, by just looking at the matrix and trying to guess some eigenvectors. In the first place, all row-sums are the same, so the vector of all 1s is an eigenvector. The associated eigenvalue is the row-sum, which is $1 + (p-1)c$. (That happens to correspond to what you claimed in your problem, but I didn't use your claim -- indeed, I didn't trust it after you changed the problem on me! -- to find the eigenvector.)
What about the other eigenvectors/values? Well, your matrix is symmetric, so eigenspaces are orthogonal, so all other eigenvectors are perpendicular to $(1, 1, \ldots, 1)$. The first vector that jumps to mind is $(1, -1, 0, 0, \ldots, 0) = e_1 - e_2$. When we multiply this by $M$, we get $(1-c)(e_1 - e_2)$, which shows that it's an eigenvector for $1-c$; it's immediately clear, having dome the multiplication by hand (I'm not going to write it out here) that $e_1 -e_3$, $e_1 - e_4$, etc., are all also eigenvectors for $1-c$. Since there are $p-1$ of these, we're done.
So my revised answer is:
- $\lambda = 1 + c(p-1)$, with multiplicity 1.
- $\lambda = 1-c$, with multiplicity $p-1$
One more alternative: you could take
$$
h(x) = det (M - xI)
$$
and solve the equation $h(x) = 0$.
Since $M = (1-c) I + Q$, where $Q$ is singular (indeed, has rank 1), it's clear that
$h(c) = 0$, indeed, that $h(c) = 0$ to order $p-1$. So
$$
h(x) = (1-c)^{p-1} (ax + b)
$$
for some unknown $a$ and $b$. The "rows sum to a constant" trick tell you that for $x_0 = 1 + c(p-1)$, we have $h(x_0) = 0$ as well, which gives us all the roots of $h$.
