I've got a little question concerning matrix operations. I am supposed to prove the following equation:

$(I + CBC^T)^{-1} = I - C(C^TC + B ^{-1})^{-1}C^T$

B and C are assumed to be invertible real square (n x n) matrices and I is the identity matrix (n x n). Also, $(C^TC + B ^{-1})^{-1}$ is assumed to exist.

Can anybody give me a hint how that's supposed to be done? I know some basic identities for transposed matrices like $(A^T)^T = A$ and $(AB)^T = B^TA^T$ but I don't know how to apply them here.

Thanks for your help!

  • $\begingroup$ Hint: the inverse of the transpose is the transpose of the inverse $\endgroup$ – Felipe Jacob Jan 28 '14 at 18:40
  • $\begingroup$ Thanks, I considered that but I still don't know what to do with the identity matrix. $\endgroup$ – IronMan12 Jan 28 '14 at 19:06

Let's go by the definition of the inverse:

$$I = (I + CBC^T)(I + CBC^T)^{-1}.$$

Let us first do some simplification:

$$I - C(C^TC + B^{-1})^{-1} C^T = I - (C^{-T} (C^TC) C^{-1} + C^{-T} B^{-1} C^{-1})^{-1} = I - (I + (CBC^T)^{-1})^{-1}.$$

So, we check:

\begin{align*} (I + CBC^T) &(I - C(C^TC + B^{-1})^{-1} C^T) = (I + CBC^T) (I - (I + (CBC^T)^{-1})^{-1}) \\ &= (I + CBC^T) - (I + CBC^T)(I + (CBC^T)^{-1})^{-1}.\tag{1} \end{align*}

It would be nice if we had:

$$(I + CBC^T)(I + (CBC^T)^{-1})^{-1} \stackrel{?}{=} CBC^T.$$

So, multiply it by $(CBC^T)^{-1}(CBC^T) = I$:

\begin{align*} (I + CBC^T)&(I + (CBC^T)^{-1})^{-1} = (I + CBC^T)(I + (CBC^T)^{-1})^{-1}(CBC^T)^{-1}(CBC^T) \\ &= (I + CBC^T) \left( (CBC^T)(I + (CBC^T)^{-1}) \right)^{-1}(CBC^T) \\ &= (I + CBC^T) \left( CBC^T + I \right)^{-1}(CBC^T) = CBC^T.\tag{2} \end{align*}

Using $(1)$ and $(2)$, we have:

\begin{align*} (I + CBC^T) &(I - C(C^TC + B^{-1})^{-1} C^T) = (I + CBC^T) - (I + CBC^T)(I + (CBC^T)^{-1})^{-1} \\ &= I + CBC^T - CBC^T = I. \end{align*}

  • $\begingroup$ Wow, very impressive. I didn't expect the prove to be that long. I thought it only consisted of a couple of basic transformations. $\endgroup$ – IronMan12 Jan 28 '14 at 21:34
  • $\begingroup$ @IronMan12 Me too. My first answer was just some basic hints, then I realized they're not enough and deleted. While I was editing it, another came, with similar hints, and it was gone by the time I undeleted this. But, who knows, maybe there is a simple trick here that we missed. $\endgroup$ – Vedran Šego Jan 28 '14 at 22:00

Just as when manipulating real number identities, you can perform the same operation to LHS and RHS preserving the identity/non-identity property if the operation you use is invertible. So, starting with $$ (I+CBC^T)^{-1}=I-C(C^TC+B^{-1})^{-1}C^T\\ $$ multiply both LHS and RHS on the left by $(I+CBC^T)$ $$ I=(I+CBC^T)(I-C(C^TC+B^{-1})^{-1}C^T)\\ $$ use now the distributive law of the product w.r.t. to sum $$ I=I+CBC^T-(I+CBC^T)C(C^TC+B^{-1})^{-1}C^T\\ $$ subtract $I+CBC^T $ from both LHS and RHS $$ -CBC^T=-(I+CBC^T)C(C^TC+B^{-1})^{-1}C^T\\ $$ multiply both LHS and RHS on the right by $(C^T)^{-1}$ $$ -CB=-(I+CBC^T)C(C^TC+B^{-1})^{-1}\\ $$ multiply both LHS and RHS on the right by $(C^TC+B^{-1})$ $$ -CB(C^TC+B^{-1})=-(I+CBC^T)C\\ $$ use the distributive law again $$ -CBC^TC-C=-C-CBC^TC\\ $$ and you finally get an identity since sum is commutative.


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