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

Question

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?

Answer 1

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}] $.