Spectral decomposition of some special matrix Let $A_{n\times n} = aI+bJ$, where $I$ is the identity matrix and $J$ is the matrix of all ones. Is it possible to find the expression of $A^{1/2}$ such that $A^{1/2}A^{1/2} = A$? In particular $A = I_{n\times n} - \frac{(1-\alpha)}{n+\alpha(2-n)}J_{n\times n}$, where $0<\alpha<1$.
 A: Here is a quick and easy way to get a matrix $A^{1/2}$, provided one is willing to limit oneself to solutions of the form
$C = \alpha I + \beta J; \tag 1$
then 
$C^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 J^2 = \alpha^2 I + 2 \alpha \beta J + \beta^2 n J = \alpha^2 I + (2 \alpha \beta + n \beta^2)J, \tag 2$
where we have used the fact that $J^2 = nJ$, which is easily verified; with
$A = a I + bJ \tag 3$
and
$C^2 = A, \tag 4$
since $I$ and $J$ are linearly independent over $\Bbb C$, we must take
$\alpha^2 = a \tag 5$
and
$2 \alpha \beta + n \beta^2 = b; \tag 6$
then
$\alpha = \pm \sqrt a, \tag 7$
and thus (6) becomes a quadratic equation in $\beta$ whose coefficients are known:
$n\beta^2 + 2 \alpha \beta - b = 0; \tag 8$
enter the quadratic formula:
$\beta = \dfrac{-2\alpha \pm \sqrt{4 \alpha^2 + 4nb}}{2n}; \tag 9$
using (5), we simplify
$\beta = \dfrac{-2\alpha \pm \sqrt{4 a + 4nb}}{2n} = \dfrac{-2\alpha \pm 2\sqrt{a + nb}}{2n} = \dfrac{-\alpha \pm I\sqrt{a + nb}}{n}; \tag{10}$
with $\alpha$ and $\beta$ given by (5) and (10) we take
$A^{1/2} = C. \tag{11}$
Note: It is not yet clear to me whether or not all solutions are of this form.  Further investigation required.  End of Note.
A: The $n\times n$ matrix $A$ has a real symmetric square root $A^{1/2}$ if and only if the eigenvalues of $A$ are nonnegative.  The eigenvalues of $J$ have been found in a number of previous Questions, namely $n$ is an eigenvalue of multiplicity one and $0$ as an eigenvalue has multiplicity $n-1$.
Therefore $A = aI+bJ$ has eigenvalues $a+bn$ with multiplicity one and $a$ with multiplicity $n-1$.
So for $A$ to be positive semi-definite (i.e. have nonnegative eigenvalues) it is required that $a \ge 0$ and $a+bn \ge 0$.  This is indeed true for the case "in particular" in the revised Question, where:
$$ a = 1\; , \; b = -\frac{(1-\alpha)}{n+\alpha(2-n)} $$
given that $0\lt \alpha \lt 1$.  When we simplify:
$$ a+bn = \frac{2\alpha}{2\alpha + n(1-\alpha)} $$
Assuming this, we can deduce the existence of positive semi-definite symmetric $A^{1/2}$ and explicitly construct one by taking an orthonormal basis of eigenvectors for $A$ to diagonalize it:
$$ D = Q^T A Q $$
where $Q$ has columns that are the orthonormalized eigenvectors and $D = \operatorname{diag}(a+bn,a,\ldots,a)$, so that:
$$ A^{1/2} = Q D^{1/2} Q^T$$
Here we mean to take square roots of the diagonal entries of $D$ to produce $D^{1/2}$, but there are additional matrices $B$ that satisfy the relationship $B^2 = A$.  For example, $B = -A^{1/2}$, but for $n\gt 1$ there are further possible matrix "square roots".
