Let $D$ be a diagonal matrix; find the inverse of $D+ab^T$

Question

It's an exercise from A Primer on Linear Models written by John F. Monahan.

Is there any special property of $ab^T$,where $a$ and $b$ are both vectors?

Answer 1

Spoiler: $ab^T$ is a rank-1 matrix. In numerical linear algebra, we call $D+ab^T$ a rank-1 update of the matrix $D$. There is a well-known formula for $(D+ab^T)^{-1}$ when $D$ is invertible:

Sherman-Morrison formula.

Do it yourself: suppose $D$ is invertible.

1. Let $x=D^{-1}a$ and $y=b$. Then $D+ab^T=D(I+xy^T)$. So, the problem boils down to finding the inverse of $I+xy^T$.
2. The inverse of any matrix, if exists, is a polynomial in that matrix (Cayley-Hamilton theorem). Hence $(I+xy)^{-1}$ is a polynomial in $I+xy^T$, and in turn a polynomial in $xy^T$.
3. Yet, for any positive integer $k$, $(xy^T)^k$ is a scalar multiple of $xy^T$. (Why?)
4. Therefore $(I+xy^T)^{-1}=pI+qxy^T$ for some scalars $p$ and $q$.
5. To find $p,q$, compare both sides of $(I+xy^T)(pI+qxy^T)=I$.
6. Now $(D+ab^T)^{-1}=(I+xy^T)^{-1}D^{-1}=(pI+qxy^T)D^{-1}$.