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

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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)

edited Jul 15 at 14:15

Key Flex

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asked Jul 15 at 13:49

Yaroslav Bulatov

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up vote
0
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favorite

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)

edited Jul 15 at 14:15

Key Flex

4,406525

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

add a commentÂ |Â

up vote
0
down vote

favorite

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)

edited Jul 15 at 14:15

Key Flex

4,406525

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

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)

edited Jul 15 at 14:15

Key Flex

4,406525

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

edited Jul 15 at 14:15

Key Flex

4,406525

edited Jul 15 at 14:15

Key Flex

4,406525

edited Jul 15 at 14:15

Key Flex

4,406525

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

asked Jul 15 at 13:49

Yaroslav Bulatov

1,82711526

add a commentÂ |Â

3 Answers
3

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oldest

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up vote
1
down vote

accepted

Let the SVD be defined as $underline overline bfG = underline overline bfU ,underline overline bfLambda ,underline overline bfV ^ + $. Then $left( underline overline bfG ^ + underline overline bfG right)^ - frac32 = underline overline bfV ,underline overline bfLambda ^ - 3underline overline bfV ^ + $ and the rest should be obvious.

EDIT. $G=ULambda V^+$. Then $(G^+G)^-3/2=VLambda ^-3V^+$

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

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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
$$A(BA)^-3/2B=(AB)^-1/2.tag*$$
If we square the LHS of (*) we get
$$A(BA)^-3/2BA(BA)^-3/2B=A(BA)^-2B=AA^-1B^-1A^-1B^-1B=B^-1A^-1=(AB)^-1.$$

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

answered Jul 15 at 14:02

Lord Shark the Unknown

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

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

EDIT. $G=ULambda V^+$. Then $(G^+G)^-3/2=VLambda ^-3V^+$

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

add a commentÂ |Â

up vote
1
down vote

accepted

EDIT. $G=ULambda V^+$. Then $(G^+G)^-3/2=VLambda ^-3V^+$

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

add a commentÂ |Â

up vote
1
down vote

accepted

EDIT. $G=ULambda V^+$. Then $(G^+G)^-3/2=VLambda ^-3V^+$

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

EDIT. $G=ULambda V^+$. Then $(G^+G)^-3/2=VLambda ^-3V^+$

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

edited Jul 15 at 21:36

loup blanc

20.4k21549

edited Jul 15 at 21:36

loup blanc

20.4k21549

edited Jul 15 at 21:36

loup blanc

20.4k21549

answered Jul 15 at 14:08

John Polcari

382111

answered Jul 15 at 14:08

John Polcari

382111

answered Jul 15 at 14:08

John Polcari

382111

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

add a commentÂ |Â

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

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.
â€“Â loup blanc
Jul 15 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...
â€“Â John Polcari
Jul 16 at 1:34

add a commentÂ |Â

up vote
1
down vote

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

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

add a commentÂ |Â

up vote
1
down vote

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

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

add a commentÂ |Â

up vote
1
down vote

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

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

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

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

answered Jul 15 at 14:02

Lord Shark the Unknown

85.8k951112

add a commentÂ |Â

up vote
1
down vote

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

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

add a commentÂ |Â

up vote
1
down vote

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

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

add a commentÂ |Â

up vote
1
down vote

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

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

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

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

edited Jul 15 at 16:23

answered Jul 15 at 16:17

greg

5,7431715

answered Jul 15 at 16:17

greg

5,7431715

answered Jul 15 at 16:17

greg

5,7431715

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

add a commentÂ |Â

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

Interesting! What are restrictions on f?
â€“Â Yaroslav Bulatov
Jul 16 at 13:23

I don't recall there being any restrictions, as long as the function can be evaluated on the two arguments.
â€“Â greg
Jul 16 at 15:56

add a commentÂ |Â

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