Solving efficiently $Ax=b$

Question

Calculating efficiently $$AX = C$$ Given $$ A = I + BB^{T} $$ where $ B $ is an orthogonal matrix of $ n \times m $.
Since $B$ is a semi-orthogonal matrix, does $ BB^{T} = I_n $ hold? If not, how do I compute $ A^{-1} $?

Answer 1

In general, $\ A^{-1} = I_n - \frac{1}{2}BB^T\ $. For $\ m \ge n\ $, as pointed out in the comments, $\ BB^T = I_n\ , A = 2I_n\ $, and the result is (almost) trivial. On the other hand, if $\ m < n\ $ then $\ B^TB = I_m\ $, and $\ BB^TBB^T = BI_mB^T = BB^T\ $. Thus \begin{eqnarray} \ A\left(I_n - \frac{1}{2}BB^T\right) &=& \left(I_n + BB^T\right)\left(I_n - \frac{1}{2}BB^T\right)\\ &=& I_n + BB^T - \frac{1}{2}BB^T - \frac{1}{2}BB^TBB^T\\ &=& I_n + BB^T - \frac{1}{2}BB^T - \frac{1}{2}BB^T \\ &=&I_n \end{eqnarray}