Finding square roots of a matrix of the form $A^\prime A$, where $A=\begin{bmatrix}I \\ a^{T}\end{bmatrix}$ for some column vector $a$
To follow up on Rahul's comment, we wish to find a matrix $M$ such that
$$M^2 = S = I + aa^T$$
Let's denote the length and direction of the $a$-vector as separate variables
$$\eqalign{
\lambda^2 &= a^Ta &\implies \lambda = \|a\| \cr
n &= \lambda^{-1}a\cr
}$$
Construct the ortho-projector $P=(I-nn^T)$ with properties
$$P^2 = P = P^T,\,\,\,\, Pn=0$$
Now consider this
$$\eqalign{
(P+\beta nn^T)^2 &= P + \beta^2 nn^T \cr
&= I + (\beta^2-1)nn^T\cr
&= I + \tfrac{(\beta^2-1)}{\lambda^2}aa^T\cr
}$$
Selecting $\,\beta={\sqrt{\lambda^2+1}}\,$ yields
$$(P+\beta nn^T)^2 = I+aa^T = S$$
Therefore
$$\eqalign{
{\sqrt S}
&= P+\beta nn^T \cr
&= I + (\beta-1)nn^T \cr
&= I + \frac{\beta-1}{\lambda^2}aa^T \cr
&= I + \frac{\beta-1}{\beta^2-1}aa^T \cr
&= I + \frac{aa^T}{1+\beta} \cr
&= I + \frac{aa^T}{1+\sqrt{a^Ta+1}} \cr\cr
}$$
The inverse can be handled using the Sherman-Morrison formula $$\eqalign{
S^{-1} &= (I+aa^T)^{-1} \cr
&= I - \frac{aa^T}{1+a^Ta} \cr\cr
}$$
Bonus result:
Any function of $\,S=(I+aa^T)\,$ can be written as
$$f(S) = \bigg(\frac{f(a^Ta+1)-f(1)}{a^Ta}\bigg)\,aa^T + f(1)\,I $$
Start with a formula (427) from the Matrix Cookbook (for idempotent $A$)
$$f(sI+tA) = (I-A)\,f(s) + A\,f(s+t)$$
and substitute $\,s=1,\,\,t=a^Ta,\,\,A=\frac{aa^T}{a^Ta}$
I don't know a name for this kind of matrix, but here is some insight on the square root, obtained from limited empirical evidence:
The eigenvalues of $S$ are $\lambda=1$ with multiplicity $n - 1$ and $\mu=1 + a_1^2 + \dots + a_n^2$ with multiplicity $1$.
The eigenvectors for $\lambda$ are $v_2=(-a_2, a_1, 0, 0, \dots, 0)$, $v_3=(-a_3, 0, a_1, 0, \dots, 0)$, $v_4=(-a_4, 0, 0, a_1, \dots, 0)$, $\dots$, $v_n=(-a_n, 0, 0, 0, \dots, a_1)$.
The eigenvectors for $\mu$ is $v_1=(a_1, \dots, a_n)$.
Then a square root for $S$ is $P \operatorname{diag}(\sqrt\mu,1,\dots,1)P^{-1}$, where $P$ is the matrix of eigenvectors.
$$A=\begin{bmatrix}I \\ a^{T}\end{bmatrix}\to\\S=A'A=A=\begin{bmatrix}I & a\end{bmatrix}\begin{bmatrix}I \\ a^{T}\end{bmatrix}=I+aa^T$$since $aa^T$ is of rank 1 it has $n-1$ zero eigenvalues and one $a^Ta$ eigenvalue therefore the eigenvalues of $S$ include $n-1$ of 1 and one $1+a^Ta$ so $S$ is invertible if and only if $$a^Ta\ne -1$$ also we have that$$S^2=(I+aa^T)(I+aa^T)=I+2aa^T+aa^Taa^T=I+2aa^T+(a^Ta)aa^T=I+(2+a^Ta)aa^T=(2+a^Ta)S-(1+a^Ta)I$$if $S$ is invertible therefore $$S=(2+a^Ta)I-(1+a^Ta)S^{-1}$$or $$S^{-1}=\dfrac{1}{1+a^Ta}\left((2+a^Ta)I-I-aa^T\right)=I-\dfrac{1}{1+a^Ta}aa^T$$now for finding the powers of $S$ let $$S^k=(1+t_k)S-t_kI$$multiplying by $S$ yields to $$S^{k+1}=(1+t_k)S^2-t_kS=(2+a^Ta+(1+a^Ta)t_k)S-(1+a^Ta)(1+t_k)I=(1+t_{k+1})S-t_{k+1}I$$therefore$$t_{k+1}=(1+a^Ta)(1+t_k)$$it's not hard to see that $$t_k=\dfrac{(1+a^Ta)^k-1-a^Ta}{a^Ta}$$and we have $$S^k=\dfrac{(1+a^Ta)^k-1}{a^Ta}S-\dfrac{(1+a^Ta)^k-1-a^Ta}{a^Ta}I$$or $$\LARGE S^n=I+\dfrac{(1+a^Ta)^n-1}{a^Ta}aa^T\quad,\quad n\in\Bbb Z$$it is interesting that the above equation holds even for negative $n$s (prove as a good exercise!). Also if $\hat{S}$ is such that $\hat{S}^2=S$ then the eigenvalues of $\hat{S}$ are $n-1$ 1 and one $\sqrt{1+a^Ta}$ which means that $$\hat{S}=I+kaa^T$$ for some k. Therefore $$\hat{S}^2=I+(k^2+2k)aa^T$$which means that $$k=\dfrac{1}{1+\sqrt{1+a^Ta}}$$and finally we obtain$$\LARGE S^{\frac{1}{2}}=I+\dfrac{1}{1+\sqrt{1+a^Ta}}aa^T$$