Why is $left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$

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While trying to generalize a certain formula from the scalar case to the matrix case, I came across the following curious observation:



Let $M$ be a real, not necessarily square matrix, then
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
which seems to be true in general. It holds, for example, for this randomly chosen case
$$M=left(
beginarrayccc
-1 & 7 & -4 \
-1 & 5 & 7 \
-4 & 7 & 4 \
-2 & -5 & 0 \
2 & -1 & -6 \
endarray
right)$$



Although numerical examples convinced me that this is true, I do not understand why. Any hints?










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  • 6




    Hint: 2/3=4/6 because 2×6=3×4
    – Empy2
    Aug 31 at 8:30














up vote
4
down vote

favorite
1












While trying to generalize a certain formula from the scalar case to the matrix case, I came across the following curious observation:



Let $M$ be a real, not necessarily square matrix, then
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
which seems to be true in general. It holds, for example, for this randomly chosen case
$$M=left(
beginarrayccc
-1 & 7 & -4 \
-1 & 5 & 7 \
-4 & 7 & 4 \
-2 & -5 & 0 \
2 & -1 & -6 \
endarray
right)$$



Although numerical examples convinced me that this is true, I do not understand why. Any hints?










share|cite|improve this question

















  • 6




    Hint: 2/3=4/6 because 2×6=3×4
    – Empy2
    Aug 31 at 8:30












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





While trying to generalize a certain formula from the scalar case to the matrix case, I came across the following curious observation:



Let $M$ be a real, not necessarily square matrix, then
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
which seems to be true in general. It holds, for example, for this randomly chosen case
$$M=left(
beginarrayccc
-1 & 7 & -4 \
-1 & 5 & 7 \
-4 & 7 & 4 \
-2 & -5 & 0 \
2 & -1 & -6 \
endarray
right)$$



Although numerical examples convinced me that this is true, I do not understand why. Any hints?










share|cite|improve this question













While trying to generalize a certain formula from the scalar case to the matrix case, I came across the following curious observation:



Let $M$ be a real, not necessarily square matrix, then
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
which seems to be true in general. It holds, for example, for this randomly chosen case
$$M=left(
beginarrayccc
-1 & 7 & -4 \
-1 & 5 & 7 \
-4 & 7 & 4 \
-2 & -5 & 0 \
2 & -1 & -6 \
endarray
right)$$



Although numerical examples convinced me that this is true, I do not understand why. Any hints?







matrices matrix-equations






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asked Aug 31 at 8:21









Wouter

5,62021435




5,62021435







  • 6




    Hint: 2/3=4/6 because 2×6=3×4
    – Empy2
    Aug 31 at 8:30












  • 6




    Hint: 2/3=4/6 because 2×6=3×4
    – Empy2
    Aug 31 at 8:30







6




6




Hint: 2/3=4/6 because 2×6=3×4
– Empy2
Aug 31 at 8:30




Hint: 2/3=4/6 because 2×6=3×4
– Empy2
Aug 31 at 8:30










2 Answers
2






active

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votes

















up vote
1
down vote



accepted










More in detail we have that if $M$ is a n-by-m matrix



  • $X=I - MM^T$ is a n-by-n matrix


  • $Y=I - M^T M$ is a m-by-m matrix


and therefore



$$X_ntimes n^-1M_ntimes m=M_ntimes mY_mtimes m^-1$$



$$X_ntimes nX_ntimes n^-1M_ntimes mY_mtimes m=X_ntimes nM_ntimes mY_mtimes m^-1Y_mtimes m$$



$$I_ntimes nM_ntimes mY_mtimes m=X_ntimes nM_ntimes mI_mtimes m$$



$$M_ntimes mY_mtimes m=X_ntimes nM_ntimes m$$



that is



$$M left(I - M^T M right) = left(I - M M^T right) M$$






share|cite|improve this answer


















  • 1




    Gimusi please fix your $X$.
    – Maam
    Aug 31 at 9:32










  • @Maam Good catch! Thanks, I fix the typo!
    – gimusi
    Aug 31 at 9:41

















up vote
3
down vote













Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$






share|cite|improve this answer






















  • Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
    – Kevin
    Aug 31 at 8:52






  • 4




    @Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
    – Arnaud D.
    Aug 31 at 10:15











  • @ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
    – Kevin
    Aug 31 at 10:18










  • @Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
    – Kevin
    Aug 31 at 10:20










  • @Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
    – Arnaud D.
    Aug 31 at 10:22










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










More in detail we have that if $M$ is a n-by-m matrix



  • $X=I - MM^T$ is a n-by-n matrix


  • $Y=I - M^T M$ is a m-by-m matrix


and therefore



$$X_ntimes n^-1M_ntimes m=M_ntimes mY_mtimes m^-1$$



$$X_ntimes nX_ntimes n^-1M_ntimes mY_mtimes m=X_ntimes nM_ntimes mY_mtimes m^-1Y_mtimes m$$



$$I_ntimes nM_ntimes mY_mtimes m=X_ntimes nM_ntimes mI_mtimes m$$



$$M_ntimes mY_mtimes m=X_ntimes nM_ntimes m$$



that is



$$M left(I - M^T M right) = left(I - M M^T right) M$$






share|cite|improve this answer


















  • 1




    Gimusi please fix your $X$.
    – Maam
    Aug 31 at 9:32










  • @Maam Good catch! Thanks, I fix the typo!
    – gimusi
    Aug 31 at 9:41














up vote
1
down vote



accepted










More in detail we have that if $M$ is a n-by-m matrix



  • $X=I - MM^T$ is a n-by-n matrix


  • $Y=I - M^T M$ is a m-by-m matrix


and therefore



$$X_ntimes n^-1M_ntimes m=M_ntimes mY_mtimes m^-1$$



$$X_ntimes nX_ntimes n^-1M_ntimes mY_mtimes m=X_ntimes nM_ntimes mY_mtimes m^-1Y_mtimes m$$



$$I_ntimes nM_ntimes mY_mtimes m=X_ntimes nM_ntimes mI_mtimes m$$



$$M_ntimes mY_mtimes m=X_ntimes nM_ntimes m$$



that is



$$M left(I - M^T M right) = left(I - M M^T right) M$$






share|cite|improve this answer


















  • 1




    Gimusi please fix your $X$.
    – Maam
    Aug 31 at 9:32










  • @Maam Good catch! Thanks, I fix the typo!
    – gimusi
    Aug 31 at 9:41












up vote
1
down vote



accepted







up vote
1
down vote



accepted






More in detail we have that if $M$ is a n-by-m matrix



  • $X=I - MM^T$ is a n-by-n matrix


  • $Y=I - M^T M$ is a m-by-m matrix


and therefore



$$X_ntimes n^-1M_ntimes m=M_ntimes mY_mtimes m^-1$$



$$X_ntimes nX_ntimes n^-1M_ntimes mY_mtimes m=X_ntimes nM_ntimes mY_mtimes m^-1Y_mtimes m$$



$$I_ntimes nM_ntimes mY_mtimes m=X_ntimes nM_ntimes mI_mtimes m$$



$$M_ntimes mY_mtimes m=X_ntimes nM_ntimes m$$



that is



$$M left(I - M^T M right) = left(I - M M^T right) M$$






share|cite|improve this answer














More in detail we have that if $M$ is a n-by-m matrix



  • $X=I - MM^T$ is a n-by-n matrix


  • $Y=I - M^T M$ is a m-by-m matrix


and therefore



$$X_ntimes n^-1M_ntimes m=M_ntimes mY_mtimes m^-1$$



$$X_ntimes nX_ntimes n^-1M_ntimes mY_mtimes m=X_ntimes nM_ntimes mY_mtimes m^-1Y_mtimes m$$



$$I_ntimes nM_ntimes mY_mtimes m=X_ntimes nM_ntimes mI_mtimes m$$



$$M_ntimes mY_mtimes m=X_ntimes nM_ntimes m$$



that is



$$M left(I - M^T M right) = left(I - M M^T right) M$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 31 at 9:41

























answered Aug 31 at 8:45









gimusi

71.7k73787




71.7k73787







  • 1




    Gimusi please fix your $X$.
    – Maam
    Aug 31 at 9:32










  • @Maam Good catch! Thanks, I fix the typo!
    – gimusi
    Aug 31 at 9:41












  • 1




    Gimusi please fix your $X$.
    – Maam
    Aug 31 at 9:32










  • @Maam Good catch! Thanks, I fix the typo!
    – gimusi
    Aug 31 at 9:41







1




1




Gimusi please fix your $X$.
– Maam
Aug 31 at 9:32




Gimusi please fix your $X$.
– Maam
Aug 31 at 9:32












@Maam Good catch! Thanks, I fix the typo!
– gimusi
Aug 31 at 9:41




@Maam Good catch! Thanks, I fix the typo!
– gimusi
Aug 31 at 9:41










up vote
3
down vote













Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$






share|cite|improve this answer






















  • Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
    – Kevin
    Aug 31 at 8:52






  • 4




    @Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
    – Arnaud D.
    Aug 31 at 10:15











  • @ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
    – Kevin
    Aug 31 at 10:18










  • @Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
    – Kevin
    Aug 31 at 10:20










  • @Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
    – Arnaud D.
    Aug 31 at 10:22














up vote
3
down vote













Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$






share|cite|improve this answer






















  • Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
    – Kevin
    Aug 31 at 8:52






  • 4




    @Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
    – Arnaud D.
    Aug 31 at 10:15











  • @ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
    – Kevin
    Aug 31 at 10:18










  • @Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
    – Kevin
    Aug 31 at 10:20










  • @Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
    – Arnaud D.
    Aug 31 at 10:22












up vote
3
down vote










up vote
3
down vote









Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$






share|cite|improve this answer














Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 31 at 10:59

























answered Aug 31 at 8:34









Wouter

5,62021435




5,62021435











  • Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
    – Kevin
    Aug 31 at 8:52






  • 4




    @Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
    – Arnaud D.
    Aug 31 at 10:15











  • @ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
    – Kevin
    Aug 31 at 10:18










  • @Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
    – Kevin
    Aug 31 at 10:20










  • @Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
    – Arnaud D.
    Aug 31 at 10:22
















  • Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
    – Kevin
    Aug 31 at 8:52






  • 4




    @Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
    – Arnaud D.
    Aug 31 at 10:15











  • @ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
    – Kevin
    Aug 31 at 10:18










  • @Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
    – Kevin
    Aug 31 at 10:20










  • @Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
    – Arnaud D.
    Aug 31 at 10:22















Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
– Kevin
Aug 31 at 8:52




Can you add this to your OP? I believe you know the rules, you have gathered extra credit for answering your own post. That in itself is fine; this extra answer need to go in your original post, and this one should be deleted.
– Kevin
Aug 31 at 8:52




4




4




@Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
– Arnaud D.
Aug 31 at 10:15





@Kevin I don't think this is necessary. It's allowed to answer one's own question, and even accept one's own answer (although it doesn't bring any reputation). See also this Meta question.
– Arnaud D.
Aug 31 at 10:15













@ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
– Kevin
Aug 31 at 10:18




@ArnaudD. Oh! I thought answering your own post in this way was against house rules. My mistake. I checked math.stackexchange.com/help/self-answer and it is totaly true. I had no idea, Thansk for your clarification and I extend my apologies to Wouter.
– Kevin
Aug 31 at 10:18












@Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
– Kevin
Aug 31 at 10:20




@Wouter My apologies, as you can see above, I am wrong. I accept that. I downvoted this anser on the basis that that you shouldn't answer your own post. I cannot re-upvote as I am locked in. I will, however, check some more of your excellent answers! All the best, Kevin
– Kevin
Aug 31 at 10:20












@Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
– Arnaud D.
Aug 31 at 10:22




@Kevin Just for your information (and perhaps Wouter's), you can retract your vote if the question is edited.
– Arnaud D.
Aug 31 at 10:22

















 

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