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Can someone suggest how this equality is derived? I feel like I'm forgetting some basic property of of matrices (this formula comes up in Figure 1 of https://arxiv.org/pdf/1806.02958.pdf)

Key Flex
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Yaroslav Bulatov
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3 Answers3

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I don't like $G$ and $G'$ as notation for matrices, so I'll write $A$ and $B$ instead. So the question is to prove $$A(BA)^{-3/2}B=(AB)^{-1/2}.\tag{*}$$ If we square the LHS of (*) we get $$A(BA)^{-3/2}BA(BA)^{-3/2}B=A(BA)^{-2}B=AA^{-1}B^{-1}A^{-1}B^{-1}B=B^{-1}A^{-1}=(AB)^{-1}.$$

So $(*)$ holds if we interpret it as saying that the expression on the left, whichever square root of $BA$ we take, when squared gives $(AB)^{-1}$.

Angina Seng
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Let the SVD be defined as $\underline {\overline {\bf{G}} } = \underline {\overline {\bf{U}} } \,\underline {\overline {\bf{\Lambda }} } \,{\underline {\overline {\bf{V}} } ^ + }$. Then ${\left( {{{\underline {\overline {\bf{G}} } }^ + }\underline {\overline {\bf{G}} } } \right)^{ - \frac{3}{2}}} = \underline {\overline {\bf{V}} } \,{\underline {\overline {\bf{\Lambda }} } ^{ - 3}}{\underline {\overline {\bf{V}} } ^ + }$ and the rest should be obvious.

EDIT. $G=U\Lambda V^+$. Then $(G^+G)^{-3/2}=V\Lambda ^{-3}V^+$

John Polcari
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  • I think that you have good ideas; unfortunately, your formulae are quasi unreadable and I think that discourages some people from reading you. That's why I'm offering you the above edit. –  Jul 15 '18 at 21:40
  • @loupblanc: Cool - I use the more tedious notation because I often have problems where there are serious mixtures of scalars, vectors, and matrices, and (to me at least) the visual ability to differentiate the three comes in handy. But it's probably not for everyone... – John Polcari Jul 16 '18 at 01:34
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Here is a wonderful result, which I attribute to Higham (but I may be wrong) $$A\,f(BA) = f(AB)\,A$$ which is true for any two matrices for which the respective function arguments are square-shaped and the functions exists.

Applying this to the current problem yields $$\eqalign{ G(G^TG)^{-3/2}G^T &= (GG^T)^{-3/2}GG^T \cr &= (GG^T)^{-1/2}(GG^T)^{-1}GG^T \cr &= (GG^T)^{-1/2} \cr }$$

greg
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