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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.

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Show that the product is the identity matrix. – Your Ad Here Feb 1 '14 at 10:07
Shouldn't it be $\;A+uv^T\color{red}I\;$ ? – DonAntonio Feb 1 '14 at 10:07
@DonAntonio: I assume $u,v$ are meant as one column, i.e. $uv^T$ is a $n\times n$ matrix. – Your Ad Here Feb 1 '14 at 10:11
yes u and v are $n\times 1$ vectors – user17981 Feb 1 '14 at 10:19
The most voted answer in here might help you. Cheers!… – Dmoreno Feb 1 '14 at 10:24
up vote 2 down vote accepted

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

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PAssing from the second to the third line isn't very clear ( to me), but anyway I think the third line's second term's numerator should have $\;A^{-1}\;$ and not $\;A\;$ ... – DonAntonio Feb 1 '14 at 11:40

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.

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