Inverse of $I+A^{T}*A$

Question

I want to calculate:

$(I-A^{T}*A)^{-1}$

and

$(I-A*A^{T})^{-1}$

Where $A$ is a square matrix and $I$ is the identity. Is there any way to develop those expression or, at least, minimize the number of operations for both expressions.

Answer 1

We need some assumptions on $A$ to give a general formula. For instance if $A$ is the identity matrix, you're asking for an inverse that doesn't exist.

If $B=A^TA$ is nilpotent, say of degree $k$, then we have $$(I-B+B^2-\dots+(-1)^{k-1} B^{k-1})(I+B)= I,$$

And deduce the inverse of $I+B$ from that equation. If $B$ is such that $\sum_{i=0}^\infty (-1)^i B^i$ converges, then $$(I-B)\sum_{i=0}^\infty(- B)^i=I.$$

If $A^TA$ is diagonalizable, then sometimes you can extract the inverse of $I-A^TA$ from the characteristic polynomial (using Cayley-Hamilton theorem).

If $A$ is orthonormal, then $I-A^TA$ is zero, therefore the inverse doesn't exist.

But in the general case, I guess you need more information.

Answer 2

We probably can't do much to compute the inverse of one of these matrices more easily. Any symmetric matrix whose eigenvalues are all at most $1$ can be expressed in the form $I - A^{\mathsf T}A$, which is fairly general.

At least the computations are not independent. The Woodbury matrix identity, which says that $$ (A+UCV)^{-1} = A^{-1} - A^{-1}U (C^{-1}+VA^{-1}U )^{-1} VA^{-1}, $$ tells us in this particular case that $$ (I - AA^{\mathsf T})^{-1} = I + A(I - A^{\mathsf T}A)^{-1}A^{\mathsf T}, $$ which can also be checked by direct multiplication. As a result, if we find $X = (I - A^{\mathsf T}A)^{-1}$, then $Y = (I - AA^{\mathsf T})^{-1}$ can be found from there without having to compute a second matrix inverse: it is given by $Y = I + AXA^{\mathsf T}$.