Show that $C=B+gh^\top$ is invertible and the inverse is given by$ D=B^{-1}-(B^{-1}gh^\top B^{-1})/(1+h^\top B^{-1}g)$

Question

My daughter set me this question and I have no idea how to solve it.It's a long time since I did any algebra.

Let $B$ be an invertible $n \times n$ matrix and $g$ and $h$ be $n$-dimensional column vectors (that is $n\times 1$ matrices) that satisfy $1+h^{T}B^{-1}g\ne0$. Show that $C=B+gh^T$ is invertible and the inverse is given by $D=B^{-1}-(B^{-1}gh^TB^{-1})/(1+h^TB^{-1}g)$ Be careful that all products that you write down/use in the process are indeed valid. Hint: What dimension is the quantity in the denominator of the formula for $D$? Hint 2: Note that for matrices $A,B,C,D$ (assuming that the dimensions are such that the products and sums are defined) we have $ABD+ACD=A(B+C)D$ (Just distribution applied twice).

This is the famous Sherman–Morrison formula that can be used to quickly update the inverse matrix after a $1$-rank perturbation. — Just a user, Oct 13 '23 at 12:38

score 1 · Answer 1 · answered Oct 12 '23 at 12:17

To show that $C$ has an inverse, you just need to compute and verify that $CD = DC = \mathbb{1}$. First compute $CD$. If you do the algebra correctly you will arrive at

$CD = (B+gh^T)(B^{-1}- \frac{1}{1+h^T B^{-1} g}B^{-1}gh^TB^{-1}) = \mathbb{1} + \frac{1}{1+h^T B^{-1} g}(h^T B^{-1} g gh^T B^{-1} - gh^TB^{-1}gh^TB^{-1}) $

Now if we prove that the last matrix is equal to $0$ we are done. To do the consider the chain of equalities

$h^TB^{-1}g g h^T B^{-1} = (h^T B^{-1} g)(gh^T)B^{-1} = g(h^T B^{-1}g)h^T B^{-1} = gh^T B^{-1} gh^T B^{-1} $

where the crucial step was to use the associativity of the matrix product and the fact that $h^T B^{-1} g \in \mathbb{R} $ is just a scalar and hence commutes with every matrix. Thus $CD = \mathbb{1}$ and $DC = \mathbb{1}$ is proved similarly.

Thank you. Very well explained. – Robert Beattie Oct 12 '23 at 23:36 — Robert Beattie, Oct 12 '23 at 23:36

score 0 · Answer 2 · answered Oct 13 '23 at 13:58

First write $$C=B(I+B^{-1}gh^\top).$$ Consequently $$ C^{-1}=(I+B^{-1}gh^\top)^{-1}B^{-1}. $$

Now explore the idea that the inverse of $I+uv^\top$ has the same structure $I-cuv^\top$, for instance from considering the formal Neumann series, and insert this guess into the defining formula of the inverse $$ (I+uv^\top)(I-cuv^\top)=I+(1-c-cv^\top u)uv^\top $$ The last terms cancel to leave the identity behind if $c=(1+v^\top u)^{-1}$.

Now apply this to the original task with $u=B^{-1}g$ and $v=h$ to get $$ C^{-1}=\left(I-(1+h^\top B^{-1}g)^{-1} B^{-1}gh^\top\right)B^{-1} = B^{-1}-\frac{B^{-1}gh^\top B^{-1}}{1+h^\top B^{-1}g} $$

Show that $C=B+gh^\top$ is invertible and the inverse is given by$ D=B^{-1}-(B^{-1}gh^\top B^{-1})/(1+h^\top B^{-1}g)$

2 Answers2