How to find eigenvalues of the following matrix? $\frac{N}{4}I_m +(\frac{N}{4}-1)J_m$,$\:$for  $N>m$, where $I_m$ is an identity matrix of size $m$ and $J_m$ is a matrix of all ones of size $m$, $N$ is a integer multiple of 4
 A: OK, first we find the eigenvalues of $J_m$, as follows: it is easy to see that
$J_m^2 = J_m J_m = mJ_m, \tag 1$
or 
$J_m^2 - mJ_m = 0; \tag 2$
(2) implies the eigenvalues of $J_m$ must lie in the set $\{0, m \}$.  It is clear that $m$ is an eigenvalue of $J_m$ with eigenvector
$\vec e_m = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}. \tag 3$
We note that 
$J_m^T = J_m, \tag{4}$
i.e., $J_m$ is symmetric; thus it is possessed of an orthogonal eigenbasis.  We proceed to construct $m - 1$ mutually orthogonal vectors corresponding to eigenvalue $0$.  Consider the vector
$\vec e_{0,1} = \begin{pmatrix} 1 - m \\ 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}; \tag 5$
we evidently have 
$J_m \vec e_{0,1} = 0, \tag 6$
that is,
$\vec e_{0,1} \in \ker J_m \tag 7$
is an eigenvector corresponding to eigenvalue $0$ of $J_m$; also, we observe that
$\vec e_m \cdot \vec e_{0,1} = \vec e_m^T \vec e_{0,1} = 0, \tag 8$
which should come as no surprise since $\vec e_m$ and $\vec e_{0,1}$ are eigenvectors of the symmetric matrix $J_m$ associated with the distinct eigenvalues $0$ and $m$.  Continuing in this vein, we define
$\vec e_{0,2} = \begin{pmatrix} 0 \\ 2 - m \\ 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}; \tag 9$
we have
$J_m \vec e_{0,2} = \vec e_m \cdot \vec e_{0,2} = \vec e_{0,1} \cdot \vec e_{0,2} = 0, \tag{10}$
and we can continue in this manner, defining $\vec e_{0i}$, until we reach $\vec e_{0,m - 1}$:
$\vec e_{0,m - 1} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ -1 \\ 1 \end{pmatrix}; \tag{11}$
at this point we have constructed $m - 1$ vectors $\vec e_{0,i}$, $1 \le i \ne j \le m - 1$, with
$J_m \vec e_{0,i} = \vec e_m \cdot \vec e_{0,i} = \vec e_{0,i} \cdot \vec e_{0,j} = 0, \tag{12}$
where $i \ne j$; we thus conclude that the multiplicity of $m$ as an eigenvalue of $J_m$ is $1$, and that of $0$ is $m - 1$.
Since the eigenvalues of $(N/4 - 1)J_m$ are are simply $N/4 - 1$ times those of $J_m$, it follows that $(N/4 - 1)J_m$ has an eigenvalue $(N/4 -1)m$ of muliplicity $1$, and $0$ of multiplicity $m - 1$.
The immediately preceding conclusion draws upon the fact that if a matrix $T$ has an eigenvalue $\mu$, so that
$T\vec v = \mu \vec v, \tag{13}$
then
$(\alpha T) \vec v = \alpha (T \vec v) = \alpha (\mu \vec v) = (\alpha \mu) \vec v \tag{14}$
i.e., $\alpha \mu$ is an eigenvalue of $\alpha T$, also with eigenvector $\vec v$.  Likewise, (13) yields
$(T + \alpha I) \vec v = T \vec v + \alpha I \vec v = \mu \vec v + \alpha \vec v = (\mu + \alpha) \vec v, \tag{15}$
which combined with out knowledge of the eigenvalues of  $(N/4 - 1)J_m$ shows that $(N/4)I_m + (N/4 - 1) J_m$ has eigenvalues
$\dfrac{N}{4} + (\dfrac{N}{4} - 1)m = \dfrac{N}{4} (m + 1) - m \tag{16}$
of multiplicity $1$ and 
$\dfrac{N}{4} \tag{17}$
of multiplicity $m - 1$.
