TITLE: 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 $a$
QUESTION [11 upvotes]: I have a matrix of quite special form:
$$S=A'A$$
$$A=\begin{bmatrix}
 1 &0  &\dots &0 \\ 
 0 &1  &\dots &0 \\ 
 0 &0  &\dots &1 \\ 
 a_{1} &a_{2} &\dots &a_{n} 
\end{bmatrix}$$
Thus $A$ is a $(n+1)\times n$ and $S$ is $n\times n$ symmetric matrix. 
I am wondering if there is some name for the matrices like $A$. 
I am implementing some numerical algorithm and I need to find inverses and square roots of very large matrices $S$. Since the matrix I am dealing with is of quite simple form, maybe there are some properties that would allow fast square root computations. 

Maybe some members of this community has some insight on how to make square root calculations faster. Does the matrix have a name I can research?

REPLY [2 votes]: $$A=\begin{bmatrix}I \\ a^{T}\end{bmatrix}\to\\S=A'A=A=\begin{bmatrix}I & a\end{bmatrix}\begin{bmatrix}I \\ a^{T}\end{bmatrix}=I+aa^T$$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$$\LARGE S^{\frac{1}{2}}=I+\dfrac{1}{1+\sqrt{1+a^Ta}}aa^T$$