# Inverse of $I_{T}+AA^{\prime }$

Suppose $I_{T}$ is a $T\times T$ identity matrix, and $A$ is a $T\times m$ matrix where $m<T.$ Is there any simple way to calculate the inverse of $% I_{T}+AA^{\prime },$ $\left( I_{T}+AA^{\prime }\right) ^{-1},$ if $% I_{T}+AA^{\prime }$ is invertible$?$ When $a$ is a vector, there is a simple way to calculate the inverse of $I_{T}+aa^{\prime }$. So, except for the Woodbury matrix identity, is there any simple way?

• What is the objection to the Woodbury matrix identity? Closely related: inverse of diagonal plus sum of rank one matrices Nov 8, 2016 at 2:36
• If the entries of $A$ are sufficiently small, then $I+AA'$ will remain positive definite and instead of calculating the inverse of that one typically prefers to solve systems $(I+AA')x = b$ by "exact" iterative methods like Conjugate Gradient. Nov 8, 2016 at 2:46
• There is no objection to the Woodbury matrix identity. Since I am dealing asymptics, I would like to have a simplest expression for the result. So I am wondering whether I can simplify the inverse of IT+AA′. Nov 8, 2016 at 2:46
• Unfortunately, the entries of A is not small, and it depends on unknown parameters. When A is a vector a, the inverse is very simple. It becomes complicated for matrix A. Nov 8, 2016 at 2:48
• There was a good Comment, now removed, that suggested the "binomial" expansion for the inverse (valid if $A$ is sufficiently small for the geometric series to converge), $(I+AA')^{-1} = \sum_{k=0}^\infty (-AA')^k$. Nov 8, 2016 at 2:50

If one is satisfied with a symbolic expression for $(I+AA')^{-1}$, a simple one can be given in terms of the singular values of $A$, or equivalently the eigenvalues of the real symmetric positive semi-definite matrix $AA'$.

Given a singular value decomposition $A = UDV'$ where $U$ is $T\times T$ orthogonal, $V$ is $m\times m$ orthogonal, and $D$ is the $T\times m$ matrix of singular values of $A$ on its diagonal:

$$D = \begin{bmatrix} \operatorname{diag}[d_1,d_2,\ldots,d_m] \\ 0 \end{bmatrix}$$

where $d_1 \ge d_2 \ge \ldots \ge d_m \ge 0$.

Note that $AA' = UDV'VD'U' = UDD'U'$, and $DD'$ is the $T\times T$ diagonal matrix:

$$DD' = \begin{bmatrix} \operatorname{diag}[d_1^2,d_2^2,\ldots,d_m^2] & 0 \\ 0 & 0 \end{bmatrix}$$

Now $I+AA' = I + UDD'U' = U(I + DD')U'$, whose inverse is expressed:

$$(I+AA')^{-1} = U (I + DD')^{-1} U'$$

because the orthogonality of $U$ means that $U^{-1} = U'$. Thus:

$$(I+AA')^{-1} = U \begin{bmatrix} E & 0 \\ 0 & I \end{bmatrix} U'$$

where $m\times m$ block $E = \operatorname{diag}[\frac{1}{1+d_1^2},\frac{1}{1+d_2^2},\ldots, \frac{1}{1+d_m^2}]$.

• Thanks a lot. This is indeed very helpful! Nov 8, 2016 at 15:13
• I'm a computational sort, so I tend to look at problems from that point of view. Here one generally needs an iterative algorithm to compute the (non-negative) eigenvalues of $AA'$ (equiv. singular values of $A$) and the orthogonal matrix $U$. But it is possible that some bounds on the inverse of $I+AA'$ can be deduced from information you have about $A$ and its dependency on a parameter. Nov 9, 2016 at 12:06