TITLE: Cholesky, Inverse, and Determinant when updating the diagonal of a symmetric positive definite matrix
QUESTION [7 upvotes]: Suppose that $A$ is a symmetric positive definite matrix and assume its dimension $n$ is large. Let $I$ be the $n \times n$ identity matrix and $m \neq 0$ be a scalar. I'm interested in computing as many of the following as possible:

det$(A + mI)$
$(A + mI)^{-1}$
$(A + mI)^{-1}B$, where $B$ is an $n \times m$ matrix
the Cholesky decomposition of $A + mI$, 

I'd like to do this for many values of $m$. However, because $n$ is large, I'd like to know if there is some update trick based on det$(A)$, $A^{-1}$, and the Cholesky decomposition of $A$. $A$ will likely not be sparse. I've been researching this for quite a while and the results haven't been all that encouraging. 
Any help, hints, suggestions, or references would be much appreciated!

REPLY [3 votes]: Use the fact that symetric positive definite matrix is similar to a diagonal matrix with positive elements. More specifically,
$$
P^{-1}AP = \operatorname{diag}\{a_1,...a_n\}
$$
where $a_i>0$ for $i \in [1, n]$, diag{$a_1,...a_n$} is diagonal matrix, and $P $ is invertible. So We have
\begin{align}
\det(A+mI)&=\det(P^{-1}(A+mI)P)
\\
&=\det(P^{-1}AP+P^{-1}(mI)P)
\\
&=\det(P^{-1}AP+mI)
\\
&=\prod \limits_{i=1}^{n}(a_i+m)
\end{align}
Since
$$
P^{-1}(A+mI)P = \operatorname{diag}(a_i+m)
$$
And
$$
\operatorname{diag}(a_i+m)^{-1}=\operatorname{diag}(\frac{1}{a_i+m})
$$
where $a_i+m\neq0$
Thus
\begin{align}
(P^{-1}(A+mI)P)^{-1}&=P(A+mI)^{-1}P^{-1}
\\
&=\operatorname{diag}(a_i+m)^{-1}
\\
&=\operatorname{diag}(\frac{1}{a_i+m})
\end{align}
And
$$(A+mI)^{-1}=P\operatorname{diag}(\frac{1}{a_i+m})P^{-1}$$