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Orthogonal decomposition with a special inner product

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0

$begingroup$

Assume that we are working on $mathbbR^p$ with an inner product induced by the positive definite matrix $mathbfG$, i.e. for $mathbff, mathbfg in mathbbR^p$ we define $langle mathbff, mathbfg rangle_mathbfG = mathbff^top mathbfG mathbfg$. This is of course a valid inner product in view of the positive-definiteness of $mathbfG$. My question is, given the null-space of another matrix $mathbfA$-let's call it $N(mathbfA)$, can we write

$$mathbff = mathbff_1 + mathbff_2,$$

with $mathbff_2 in N(mathbfA), f_1 in N(mathbfA)^perp$ and $langle mathbff_1, mathbff_2 rangle_mathbfG = 0$? In other words, is this special but important case covered by the projection theorem? Thank you very much for your help.

linear-algebra hilbert-spaces projection

share|cite|improve this question

edited 2 days ago

JohnK

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take $mathbff_1=0$
  $endgroup$
  – Bertrand
  Mar 5 at 13:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Will not do.....
  $endgroup$
  – JohnK
  Mar 5 at 14:02
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume that we are working on $mathbbR^p$ with an inner product induced by the positive definite matrix $mathbfG$, i.e. for $mathbff, mathbfg in mathbbR^p$ we define $langle mathbff, mathbfg rangle_mathbfG = mathbff^top mathbfG mathbfg$. This is of course a valid inner product in view of the positive-definiteness of $mathbfG$. My question is, given the null-space of another matrix $mathbfA$-let's call it $N(mathbfA)$, can we write

$$mathbff = mathbff_1 + mathbff_2,$$

with $mathbff_2 in N(mathbfA), f_1 in N(mathbfA)^perp$ and $langle mathbff_1, mathbff_2 rangle_mathbfG = 0$? In other words, is this special but important case covered by the projection theorem? Thank you very much for your help.

linear-algebra hilbert-spaces projection

share|cite|improve this question

edited 2 days ago

JohnK

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take $mathbff_1=0$
  $endgroup$
  – Bertrand
  Mar 5 at 13:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Will not do.....
  $endgroup$
  – JohnK
  Mar 5 at 14:02
  

  
  
  
  

add a comment | 

$begingroup$

Assume that we are working on $mathbbR^p$ with an inner product induced by the positive definite matrix $mathbfG$, i.e. for $mathbff, mathbfg in mathbbR^p$ we define $langle mathbff, mathbfg rangle_mathbfG = mathbff^top mathbfG mathbfg$. This is of course a valid inner product in view of the positive-definiteness of $mathbfG$. My question is, given the null-space of another matrix $mathbfA$-let's call it $N(mathbfA)$, can we write

$$mathbff = mathbff_1 + mathbff_2,$$

with $mathbff_2 in N(mathbfA), f_1 in N(mathbfA)^perp$ and $langle mathbff_1, mathbff_2 rangle_mathbfG = 0$? In other words, is this special but important case covered by the projection theorem? Thank you very much for your help.

linear-algebra hilbert-spaces projection

share|cite|improve this question

edited 2 days ago

JohnK

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

$endgroup$

share|cite|improve this question

edited 2 days ago

JohnK

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

share|cite|improve this question

edited 2 days ago

JohnK

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

asked Mar 5 at 11:18

JohnKJohnK

2,84711739

1 Answer
1

active

oldest

votes

2

$begingroup$

Yes, there is always such a decomposition, and we don't even need to refer to a particular matrix $bf A$. Every null space of a matrix is a subspace, and every subspace is the null space of some matrix, so we may as well just show the claim for subspaces $S$.

For any inner product $langle ,cdot,,,cdot,rangle$ on $Bbb R^p$, any subspace $S subset Bbb R^p$ defines an orthogonal subspace $$S^perp := bf x in S : langle bf x, bf y rangle = 0 textrm for all $bf y in S$ .$$ Now, nondegeneracy implies that $dim S + dim S^perp = p$, and positive definiteness implies that $S cap S^perp = bf 0$: For any $bf x neq bf 0$ we have $langle bf x, bf x rangle > 0$, so if $bf x in S$, we must have $bf x notin S^perp$.

Thus, $Bbb R^p$ decomposes as an orthogonal direct sum $$boxedBbb R^p = S oplus S^perp .$$
In particular, we can decompose any element $bf x in Bbb R^p$ uniquely as a sum $$bf x = bf x^top + bf x^perp, qquad textrmwhere qquad bf x^top in S, bf x^perp in S^perp ,$$ and by definition $langle bf x^top, bf x^perp rangle = 0$.

share|cite|improve this answer

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the clarification.
  $endgroup$
  – JohnK
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I hope you found it useful!
  $endgroup$
  – Travis
  yesterday
  

  
  
  
  

add a comment | 

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2

$begingroup$

Yes, there is always such a decomposition, and we don't even need to refer to a particular matrix $bf A$. Every null space of a matrix is a subspace, and every subspace is the null space of some matrix, so we may as well just show the claim for subspaces $S$.

For any inner product $langle ,cdot,,,cdot,rangle$ on $Bbb R^p$, any subspace $S subset Bbb R^p$ defines an orthogonal subspace $$S^perp := bf x in S : langle bf x, bf y rangle = 0 textrm for all $bf y in S$ .$$ Now, nondegeneracy implies that $dim S + dim S^perp = p$, and positive definiteness implies that $S cap S^perp = bf 0$: For any $bf x neq bf 0$ we have $langle bf x, bf x rangle > 0$, so if $bf x in S$, we must have $bf x notin S^perp$.

Thus, $Bbb R^p$ decomposes as an orthogonal direct sum $$boxedBbb R^p = S oplus S^perp .$$
In particular, we can decompose any element $bf x in Bbb R^p$ uniquely as a sum $$bf x = bf x^top + bf x^perp, qquad textrmwhere qquad bf x^top in S, bf x^perp in S^perp ,$$ and by definition $langle bf x^top, bf x^perp rangle = 0$.

share|cite|improve this answer

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the clarification.
  $endgroup$
  – JohnK
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I hope you found it useful!
  $endgroup$
  – Travis
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, there is always such a decomposition, and we don't even need to refer to a particular matrix $bf A$. Every null space of a matrix is a subspace, and every subspace is the null space of some matrix, so we may as well just show the claim for subspaces $S$.

For any inner product $langle ,cdot,,,cdot,rangle$ on $Bbb R^p$, any subspace $S subset Bbb R^p$ defines an orthogonal subspace $$S^perp := bf x in S : langle bf x, bf y rangle = 0 textrm for all $bf y in S$ .$$ Now, nondegeneracy implies that $dim S + dim S^perp = p$, and positive definiteness implies that $S cap S^perp = bf 0$: For any $bf x neq bf 0$ we have $langle bf x, bf x rangle > 0$, so if $bf x in S$, we must have $bf x notin S^perp$.

Thus, $Bbb R^p$ decomposes as an orthogonal direct sum $$boxedBbb R^p = S oplus S^perp .$$
In particular, we can decompose any element $bf x in Bbb R^p$ uniquely as a sum $$bf x = bf x^top + bf x^perp, qquad textrmwhere qquad bf x^top in S, bf x^perp in S^perp ,$$ and by definition $langle bf x^top, bf x^perp rangle = 0$.

share|cite|improve this answer

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for the clarification.
  $endgroup$
  – JohnK
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I hope you found it useful!
  $endgroup$
  – Travis
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Yes, there is always such a decomposition, and we don't even need to refer to a particular matrix $bf A$. Every null space of a matrix is a subspace, and every subspace is the null space of some matrix, so we may as well just show the claim for subspaces $S$.

For any inner product $langle ,cdot,,,cdot,rangle$ on $Bbb R^p$, any subspace $S subset Bbb R^p$ defines an orthogonal subspace $$S^perp := bf x in S : langle bf x, bf y rangle = 0 textrm for all $bf y in S$ .$$ Now, nondegeneracy implies that $dim S + dim S^perp = p$, and positive definiteness implies that $S cap S^perp = bf 0$: For any $bf x neq bf 0$ we have $langle bf x, bf x rangle > 0$, so if $bf x in S$, we must have $bf x notin S^perp$.

Thus, $Bbb R^p$ decomposes as an orthogonal direct sum $$boxedBbb R^p = S oplus S^perp .$$
In particular, we can decompose any element $bf x in Bbb R^p$ uniquely as a sum $$bf x = bf x^top + bf x^perp, qquad textrmwhere qquad bf x^top in S, bf x^perp in S^perp ,$$ and by definition $langle bf x^top, bf x^perp rangle = 0$.

share|cite|improve this answer

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

$endgroup$

share|cite|improve this answer

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

edited 19 hours ago

Bertrand

45815

edited 19 hours ago

Bertrand

45815

answered 2 days ago

TravisTravis

62.6k767149

answered 2 days ago

TravisTravis

62.6k767149