How to invert sum of matrices?

Question

Given are two matrices:

$\bf A, \bf B$

We know that matrices $\bf A \neq \bf B$ are invertable, symmetric, positive-definite and of full rank. Is it possible to give the formula for following sum of these matrices:

$[\bf A + \lambda\bf B]^{-1} = ?$

where $\lambda$ is a scalar such as $0 < \lambda < 1$.

It is possible for $A$ and $B$ to be invertible without $A+\lambda B$ being so. For instance take $A=B$ and $\lambda =-1$. — Git Gud, Mar 02 '13 at 20:23
Still the same problem, you could have $B=-\frac{1}{\lambda}A$. — Integral, Mar 02 '13 at 20:28
Ok, let's assume that $A + \lambda B$ is invertible. If it's not, then it's irrelevant anyway. — Josephine Moeller, Mar 02 '13 at 20:29
In fact, if $A$ is positive-definite, $-1\frac{1}{\lambda}A$ isn't positive-definite. Maybe the contiditions on $\lambda$ are enough. — Integral, Mar 02 '13 at 20:39
The space of positive definite matrices is convex. So $A + \lambda B$ is invertible for $\lambda > 0$. — achille hui, Mar 02 '13 at 20:40

score 11 · Accepted Answer · edited Mar 03 '13 at 19:47

11

Assuming that $A+\lambda B$ is also invertible, you can use the Binomial Inverse Theorem:

$[A+\lambda B]^{-1} = A^{-1}-\lambda A^{-1}(I + \lambda BA^{-1})^{-1}BA^{-1}$

edited Mar 03 '13 at 19:47

Gacek

133

answered Mar 02 '13 at 20:38

Josephine Moeller

2,601

score 3 · Answer 2 · answered Mar 02 '13 at 20:48

3

Use Woodbury matrix identity:

$$ \left(A+UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1}+VA^{-1}U \right)^{-1} VA^{-1}. $$

answered Mar 02 '13 at 20:48

Elias Costa

14,658

How to invert sum of matrices?

2 Answers2

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