Inverse square root of a matrix with specific pattern Consider the following n-dimensional square (covariance) matrix:
$$\Sigma = \mathbb{1_n}\mathbb{1_n}^T + diag(\mathbb{1_n}),$$
(i.e., a matrix of one's but with a diagonal of two's).
I'm trying to find its inverse square root, $\Sigma^{-1/2}$. Given its simple form, I'm hoping to find some simple formula which would depend on $n$. So far, my attempts were unsuccessful... I would greatly appreciate your input on this. 
I'm also looking for some sort of "bible" which would gather such linear algebra identities. Would you have any good book in mind? 
Thanks!
 A: $\Sigma$ has the eigenvalue $n+1$ with algebraic and geometric multiplicity $1$ and the eigenvalue $1$ with algebraic and geometric multiplicity $n-1$. The normalized eigenvector associated with the eigenvalue $n+1$ is $\frac{1}{\sqrt{n}}1_n$. As $\Sigma$ is symmetric, we have an orthogonal matrix $S$ with 
$$
\Sigma
= S\begin{pmatrix} n+1 & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1 \end{pmatrix} S^T
$$
where the first column of $S$ is $\frac{1}{\sqrt{n}}1_n$.
This results in
$$
\Sigma^p
= S\begin{pmatrix} (n+1)^p & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1 \end{pmatrix} S^T
\\
= S\left[ \begin{pmatrix} (n+1)^p-1 & & & \\ & 0 & \dots & 0 \\ & \vdots & \ddots & \vdots \\ & 0 & \dots & 0 \end{pmatrix} +I_n \right] S^T
\\
= S\begin{pmatrix} (n+1)^p-1 & & & \\ & 0 & \dots & 0 \\ & \vdots & \ddots & \vdots \\ & 0 & \dots & 0 \end{pmatrix} S^T + I_n
$$
As we know the first column of $S$, we can simplify further:
$$\Sigma^p = 
\frac{1}{\sqrt{n}}1_n \left((n+1)^p-1\right) \frac{1}{\sqrt{n}}1_n^T + I_n
=\frac{(n+1)^p-1}{n} \, 1_n1_n^T +I_n
$$
