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?
matrices matrix-equations
add a comment |Â
up vote
4
down vote
favorite
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
6
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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
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
matrices matrix-equations
asked Aug 31 at 8:21
Wouter
5,62021435
5,62021435
6
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30
add a comment |Â
6
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30
6
6
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30
add a comment |Â
2 Answers
2
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
$$M left(I - M^T M right) = left(I - M M^T right) M$$
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
add a comment |Â
up vote
3
down vote
Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$
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.
â 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.
â 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.
â 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
â 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.
â Arnaud D.
Aug 31 at 10:22
add a comment |Â
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
$$M left(I - M^T M right) = left(I - M M^T right) M$$
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
add a comment |Â
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
$$M left(I - M^T M right) = left(I - M M^T right) M$$
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
add a comment |Â
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
$$M left(I - M^T M right) = left(I - M M^T right) M$$
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
$$M left(I - M^T M right) = left(I - M M^T right) M$$
edited Aug 31 at 9:41
answered Aug 31 at 8:45
gimusi
71.7k73787
71.7k73787
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
add a comment |Â
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
1
1
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
Gimusi please fix your $X$.
â Maam
Aug 31 at 9:32
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
@Maam Good catch! Thanks, I fix the typo!
â gimusi
Aug 31 at 9:41
add a comment |Â
up vote
3
down vote
Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$
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.
â 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.
â 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.
â 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
â 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.
â Arnaud D.
Aug 31 at 10:22
add a comment |Â
up vote
3
down vote
Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$
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.
â 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.
â 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.
â 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
â 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.
â Arnaud D.
Aug 31 at 10:22
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$
Thanks to @Empy2, just multiply both sides with the "denominator" from the other side:
$$left(I - M M^T right)^-1 M = Mleft(I - M^T M right)^-1$$
$$iff M left(I - M^T M right) = left(I - M M^T right) M$$
$$iff M - M M^T M = M - M M^T M$$
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.
â 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.
â 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.
â 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
â 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.
â Arnaud D.
Aug 31 at 10:22
add a comment |Â
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.
â 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.
â 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.
â 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
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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
â 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
â 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.
â 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.
â Arnaud D.
Aug 31 at 10:22
add a comment |Â
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6
Hint: 2/3=4/6 because 2Ã6=3Ã4
â Empy2
Aug 31 at 8:30