Proof of matrix inverse

Question

Prove that $(A+uv^T)^{-1}=A^{-1}-\frac{1}{1+v^TA^{-1}u}A^{-1}uv^TA^{-1}$

Can someone give a hint how to show it.I am not getting from where to start.

Answer 1

\begin{align} &(A+uv^T) \left(A^{-1}-\frac{1}{1+v^TA^{-1}u}A^{-1}uv^TA^{-1}\right)=\\ &=I-\frac{1}{1+v^TA^{-1}u}uv^TA^{-1}+ uv^TA^{-1}-\frac{1}{1+v^TA^{-1}u}uv^TA^{-1}uv^TA^{-1}\\ &=I+\frac{v^TAu}{1+v^TA^{-1}u}uv^TA^{-1}-\frac{v^TA^{-1}u}{1+v^TA^{-1}u}uv^TA^{-1}=I, \end{align} since $$ uv^TA^{-1}uv^TA^{-1}=(v^TA^{-1}u)uv^TA^{-1}, $$ as $v^TA^{-1}u$ is scalar. Thus $$ (A+uv^T)^{-1}=A^{-1}-\frac{1}{1+v^TA^{-1}u}A^{-1}uv^TA^{-1}. $$

Answer 2

HINT. Just calculate the product of the RHS and $A+uv^T$, and use that $v^TA^{-1}u$ is a scalar and hence commutes with everything.