proving $G(G'G)^{-3/2}G'=(GG')^{-1/2}$

Question

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)

Answer 1

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

Answer 2

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 $(AB)^{-1}$.

Answer 3

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 $$\eqalign{ G(G^TG)^{-3/2}G^T &= (GG^T)^{-3/2}GG^T \cr &= (GG^T)^{-1/2}(GG^T)^{-1}GG^T \cr &= (GG^T)^{-1/2} \cr }$$