$$(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$$
I took the inverse of both sides.
$$((C+DD^{T})^{-1})^{-1}=(C^{-1}D(I+D^{T}C^{-1}D)^{-1})^{-1}$$ $$C+DD^{T}=(I+D^{T}C^{-1}D)(C^{-1}D)^{-1}$$
After this I got stuck, what else can I do?
Show that $(C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}$
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linear-algebra
matrices
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0Do you already have the fact that $(AB)^{-1} = B^{-1}A^{-1}$? – 2017-02-14
3 Answers
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In order to verify whether $A^{-1}=B$, it's sufficient to verify that $AB=I$ (the identity matrix). Now $$ (C+DD^{T})C^{-1}D(I+D^{T}C^{-1}D)^{-1}=I $$ if and only if $$ (C+DD^{T})C^{-1}D=I+D^{T}C^{-1}D $$ that is, $$ D+DD^TC^{-1}D=I+D^TC^{-1}D $$ which is obviously false, in general. Just take $C=I$ and $D=2I$: the left hand side is $10I$, the right hand side is $5I$.
If you want to prove instead that $$ (C+DD^{T})^{-1}=C^{-1}D(I+D^{T}C^{-1}D)^{-1}\color{red}{D^{-1}} $$ this is equivalent to $$ (C+DD^{T})C^{-1}D(I+D^{T}C^{-1}D)^{-1}D^{-1}=I $$ that is equivalent to $$ (C+DD^{T})C^{-1}D(I+D^{T}C^{-1}D)^{-1}=D $$ hence to $$ (C+DD^{T})C^{-1}D=D(I+D^{T}C^{-1}D) $$ which is obviously true.
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The identity should be $$ (C+DD^T)^{-1} = C^{-1} D (I+D^T C^{-1} D)^{-1} D^{-1} $$ (otherwise the powers of $D$ don't make sense for the first term in the bracket, for one thing)
Starting on the left, $$\begin{align} (C+DD^T)^{-1} &= ((I+DD^TC^{-1})C)^{-1} \\ &= C^{-1} (I+DD^T C^{-1})^{-1} \\ &= C^{-1} DD^{-1}(I+DD^T C^{-1})^{-1} \\ &= C^{-1}D (D+DD^T C^{-1}D)^{-1} \\ &= C^{-1}D (D(I+D^T C^{-1} D))^{-1} \\ &= C^{-1} D (I+D^T C^{-1} D)^{-1} D^{-1}. \end{align}$$
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$C+DD^{T}=(I+D^{T}C^{-1}D)(C^{-1}D)^{-1} \implies C+DD^{T}=(I+D^{T}C^{-1}D)(D^{-1}C) \implies C+DD^{T}= D^{-1}C + D^{T}$.
Your identity is missing some thing.