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.

@DonAntonio: I assume $u,v$ are meant as one column, i.e. $uv^T$ is a $n\times n$ matrix. — J.R., Feb 01 '14 at 10:11
The most voted answer in here might help you.
Cheers!

http://math.stackexchange.com/questions/17776/inverse-of-the-sum-of-matrices — Dmoreno, Feb 01 '14 at 10:24

score 2 · Answer 1 · answered Feb 01 '14 at 10:43

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.

answered Feb 01 '14 at 10:43

Carsten S

8,726

score 2 · Accepted Answer · answered Feb 01 '14 at 10:56

2

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

answered Feb 01 '14 at 10:56

Yiorgos S. Smyrlis

83,933

1

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 01 '14 at 11:40

Proof of matrix inverse

2 Answers2