$A in mathbb{C}^{mtimes n}$,$A=FG^*$ and $r(A)=r(F)=r(G)$. Prove $A^dagger = G(F^*AG)^{-1}F^*$ and $A^dagger...












5












$begingroup$


Let $A^dagger$ be a Moore-Penrose inverse of a matrix $A$.



If $A in mathbb{C}^{mtimes n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that
$$A^dagger = G(F^*AG)^{-1}F^*$$ and $$A^dagger = (G^dagger)^*F^dagger.$$



I need to show this using SVD decomposition and maybe some other properties of a Moore-Penrose inverse.



I tried to show the statement by writing SVD decomposition of all the matrices included, but it just gets messy and I didn't succeed.



Any hints would be really helpful! Thanks in advance!






linear-algebra svd generalized-inverse







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do we know the sizes of $F$ and $G$?
    $endgroup$
    – Omnomnomnom
    Jan 6 at 14:11










  • $begingroup$
    @Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
    $endgroup$
    – mathbbandstuff
    Jan 6 at 14:31


















5












$begingroup$


Let $A^dagger$ be a Moore-Penrose inverse of a matrix $A$.



If $A in mathbb{C}^{mtimes n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that
$$A^dagger = G(F^*AG)^{-1}F^*$$ and $$A^dagger = (G^dagger)^*F^dagger.$$



I need to show this using SVD decomposition and maybe some other properties of a Moore-Penrose inverse.



I tried to show the statement by writing SVD decomposition of all the matrices included, but it just gets messy and I didn't succeed.



Any hints would be really helpful! Thanks in advance!






linear-algebra svd generalized-inverse







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do we know the sizes of $F$ and $G$?
    $endgroup$
    – Omnomnomnom
    Jan 6 at 14:11










  • $begingroup$
    @Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
    $endgroup$
    – mathbbandstuff
    Jan 6 at 14:31
















5












5








5


1



$begingroup$


Let $A^dagger$ be a Moore-Penrose inverse of a matrix $A$.



If $A in mathbb{C}^{mtimes n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that
$$A^dagger = G(F^*AG)^{-1}F^*$$ and $$A^dagger = (G^dagger)^*F^dagger.$$



I need to show this using SVD decomposition and maybe some other properties of a Moore-Penrose inverse.



I tried to show the statement by writing SVD decomposition of all the matrices included, but it just gets messy and I didn't succeed.



Any hints would be really helpful! Thanks in advance!






linear-algebra svd generalized-inverse







share|cite|improve this question









$endgroup$




Let $A^dagger$ be a Moore-Penrose inverse of a matrix $A$.



If $A in mathbb{C}^{mtimes n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that
$$A^dagger = G(F^*AG)^{-1}F^*$$ and $$A^dagger = (G^dagger)^*F^dagger.$$



I need to show this using SVD decomposition and maybe some other properties of a Moore-Penrose inverse.



I tried to show the statement by writing SVD decomposition of all the matrices included, but it just gets messy and I didn't succeed.



Any hints would be really helpful! Thanks in advance!






linear-algebra svd generalized-inverse




linear-algebra svd generalized-inverse






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 13:44









mathbbandstuffmathbbandstuff

283111




283111












  • $begingroup$
    Do we know the sizes of $F$ and $G$?
    $endgroup$
    – Omnomnomnom
    Jan 6 at 14:11










  • $begingroup$
    @Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
    $endgroup$
    – mathbbandstuff
    Jan 6 at 14:31




















  • $begingroup$
    Do we know the sizes of $F$ and $G$?
    $endgroup$
    – Omnomnomnom
    Jan 6 at 14:11










  • $begingroup$
    @Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
    $endgroup$
    – mathbbandstuff
    Jan 6 at 14:31


















$begingroup$
Do we know the sizes of $F$ and $G$?
$endgroup$
– Omnomnomnom
Jan 6 at 14:11




$begingroup$
Do we know the sizes of $F$ and $G$?
$endgroup$
– Omnomnomnom
Jan 6 at 14:11












$begingroup$
@Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
$endgroup$
– mathbbandstuff
Jan 6 at 14:31






$begingroup$
@Omnomnomnom No, they can be of any size, $F in mathbb{C}^{m times p}$ and $G in mathbb{C}^{n times p}$
$endgroup$
– mathbbandstuff
Jan 6 at 14:31












1 Answer
1






active

oldest

votes


















3












$begingroup$

Both equalities are technically false. The first one is false because in general, $F^ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.



However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^ast AG)^{-1}=(F^ast FG^ast G)^{-1}=(G^ast G)^{-1}(F^ast F)^{-1}$ and $F^dagger F=G^dagger G=I$. Now you can easily prove that $A^dagger=G(G^ast G)^{-1}(F^ast F)^{-1}F^ast=(G^dagger)^ast F^dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
    $endgroup$
    – mathbbandstuff
    Jan 7 at 13:24








  • 1




    $begingroup$
    @mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
    $endgroup$
    – user1551
    Jan 7 at 14:17










  • $begingroup$
    Thank you very much!
    $endgroup$
    – mathbbandstuff
    Jan 7 at 14:31











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Both equalities are technically false. The first one is false because in general, $F^ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.



However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^ast AG)^{-1}=(F^ast FG^ast G)^{-1}=(G^ast G)^{-1}(F^ast F)^{-1}$ and $F^dagger F=G^dagger G=I$. Now you can easily prove that $A^dagger=G(G^ast G)^{-1}(F^ast F)^{-1}F^ast=(G^dagger)^ast F^dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
    $endgroup$
    – mathbbandstuff
    Jan 7 at 13:24








  • 1




    $begingroup$
    @mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
    $endgroup$
    – user1551
    Jan 7 at 14:17










  • $begingroup$
    Thank you very much!
    $endgroup$
    – mathbbandstuff
    Jan 7 at 14:31
















3












$begingroup$

Both equalities are technically false. The first one is false because in general, $F^ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.



However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^ast AG)^{-1}=(F^ast FG^ast G)^{-1}=(G^ast G)^{-1}(F^ast F)^{-1}$ and $F^dagger F=G^dagger G=I$. Now you can easily prove that $A^dagger=G(G^ast G)^{-1}(F^ast F)^{-1}F^ast=(G^dagger)^ast F^dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
    $endgroup$
    – mathbbandstuff
    Jan 7 at 13:24








  • 1




    $begingroup$
    @mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
    $endgroup$
    – user1551
    Jan 7 at 14:17










  • $begingroup$
    Thank you very much!
    $endgroup$
    – mathbbandstuff
    Jan 7 at 14:31














3












3








3





$begingroup$

Both equalities are technically false. The first one is false because in general, $F^ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.



However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^ast AG)^{-1}=(F^ast FG^ast G)^{-1}=(G^ast G)^{-1}(F^ast F)^{-1}$ and $F^dagger F=G^dagger G=I$. Now you can easily prove that $A^dagger=G(G^ast G)^{-1}(F^ast F)^{-1}F^ast=(G^dagger)^ast F^dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.






share|cite|improve this answer











$endgroup$



Both equalities are technically false. The first one is false because in general, $F^ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.



However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^ast AG)^{-1}=(F^ast FG^ast G)^{-1}=(G^ast G)^{-1}(F^ast F)^{-1}$ and $F^dagger F=G^dagger G=I$. Now you can easily prove that $A^dagger=G(G^ast G)^{-1}(F^ast F)^{-1}F^ast=(G^dagger)^ast F^dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 14:12

























answered Jan 6 at 19:01









user1551user1551

72.2k566127




72.2k566127








  • 1




    $begingroup$
    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
    $endgroup$
    – mathbbandstuff
    Jan 7 at 13:24








  • 1




    $begingroup$
    @mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
    $endgroup$
    – user1551
    Jan 7 at 14:17










  • $begingroup$
    Thank you very much!
    $endgroup$
    – mathbbandstuff
    Jan 7 at 14:31














  • 1




    $begingroup$
    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
    $endgroup$
    – mathbbandstuff
    Jan 7 at 13:24








  • 1




    $begingroup$
    @mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
    $endgroup$
    – user1551
    Jan 7 at 14:17










  • $begingroup$
    Thank you very much!
    $endgroup$
    – mathbbandstuff
    Jan 7 at 14:31








1




1




$begingroup$
Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
$endgroup$
– mathbbandstuff
Jan 7 at 13:24






$begingroup$
Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^*AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^*AG$ is always well-defined because of $A=FG^*$.
$endgroup$
– mathbbandstuff
Jan 7 at 13:24






1




1




$begingroup$
@mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
$endgroup$
– user1551
Jan 7 at 14:17




$begingroup$
@mathbbandstuff You are right that $F^ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^ast AG=1$ is nonsingular when $F=G=pmatrix{1\ 0}$, but $F^ast AG=pmatrix{1&0\ 0&0}$ is singular when $F=G=pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices.
$endgroup$
– user1551
Jan 7 at 14:17












$begingroup$
Thank you very much!
$endgroup$
– mathbbandstuff
Jan 7 at 14:31




$begingroup$
Thank you very much!
$endgroup$
– mathbbandstuff
Jan 7 at 14:31


















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