Finding $limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I) ^{-1} mathbf A^T$
$begingroup$
I am given a problem to find$$lim_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1}mathbf A^T.$$
It is fairly obvious that it reduces to $(mathbf A^Tmathbf A)^{-1}mathbf A^T$, but I am not sure I can say anything meaningful beyond that. I tried plugging in some values to MATLAB, just to see if I can notice a pattern, but other than returning some "close to singular" errors when I take the inverse of a matrix, I do not see anything that stands out.
Is there something obvious I am missing here?
linear-algebra limits
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
$endgroup$
add a comment |
$begingroup$
I am given a problem to find$$lim_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1}mathbf A^T.$$
It is fairly obvious that it reduces to $(mathbf A^Tmathbf A)^{-1}mathbf A^T$, but I am not sure I can say anything meaningful beyond that. I tried plugging in some values to MATLAB, just to see if I can notice a pattern, but other than returning some "close to singular" errors when I take the inverse of a matrix, I do not see anything that stands out.
Is there something obvious I am missing here?
linear-algebra limits
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
$endgroup$
$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
2
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12
add a comment |
$begingroup$
I am given a problem to find$$lim_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1}mathbf A^T.$$
It is fairly obvious that it reduces to $(mathbf A^Tmathbf A)^{-1}mathbf A^T$, but I am not sure I can say anything meaningful beyond that. I tried plugging in some values to MATLAB, just to see if I can notice a pattern, but other than returning some "close to singular" errors when I take the inverse of a matrix, I do not see anything that stands out.
Is there something obvious I am missing here?
linear-algebra limits
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
$endgroup$
I am given a problem to find$$lim_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1}mathbf A^T.$$
It is fairly obvious that it reduces to $(mathbf A^Tmathbf A)^{-1}mathbf A^T$, but I am not sure I can say anything meaningful beyond that. I tried plugging in some values to MATLAB, just to see if I can notice a pattern, but other than returning some "close to singular" errors when I take the inverse of a matrix, I do not see anything that stands out.
Is there something obvious I am missing here?
linear-algebra limits
linear-algebra limits
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
edited Dec 6 '18 at 6:44
W. MacTurk
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
asked Dec 6 '18 at 2:40
W. MacTurkW. MacTurk
105
105
$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
2
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12
add a comment |
$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
2
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12
$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
2
2
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think I have it.
We have
$limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1} mathbf A^T$
Which reduces to
$(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$
Under SVD, this becomes
$(textbf{V} Sigma^{2} textbf{V}^{T})^{-1}textbf{A}^{T}$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{A}^T$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{V}Sigmatextbf{U}^T$
$textbf{V}Sigma^{-1}textbf{U}^T$
$= (textbf{A}^T){^{-1}}$
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
$endgroup$
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
|
show 6 more comments
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1 Answer
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1 Answer
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$begingroup$
I think I have it.
We have
$limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1} mathbf A^T$
Which reduces to
$(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$
Under SVD, this becomes
$(textbf{V} Sigma^{2} textbf{V}^{T})^{-1}textbf{A}^{T}$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{A}^T$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{V}Sigmatextbf{U}^T$
$textbf{V}Sigma^{-1}textbf{U}^T$
$= (textbf{A}^T){^{-1}}$
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
$endgroup$
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
|
show 6 more comments
$begingroup$
I think I have it.
We have
$limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1} mathbf A^T$
Which reduces to
$(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$
Under SVD, this becomes
$(textbf{V} Sigma^{2} textbf{V}^{T})^{-1}textbf{A}^{T}$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{A}^T$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{V}Sigmatextbf{U}^T$
$textbf{V}Sigma^{-1}textbf{U}^T$
$= (textbf{A}^T){^{-1}}$
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
$endgroup$
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
|
show 6 more comments
$begingroup$
I think I have it.
We have
$limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1} mathbf A^T$
Which reduces to
$(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$
Under SVD, this becomes
$(textbf{V} Sigma^{2} textbf{V}^{T})^{-1}textbf{A}^{T}$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{A}^T$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{V}Sigmatextbf{U}^T$
$textbf{V}Sigma^{-1}textbf{U}^T$
$= (textbf{A}^T){^{-1}}$
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
$endgroup$
I think I have it.
We have
$limlimits_{ε→0}(mathbf A^Tmathbf A+εmathbf I)^{-1} mathbf A^T$
Which reduces to
$(textbf{A}^Ttextbf{A})^{-1}textbf{A}^T$
Under SVD, this becomes
$(textbf{V} Sigma^{2} textbf{V}^{T})^{-1}textbf{A}^{T}$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{A}^T$
$(textbf{V}Sigma^{-1}Sigma^{-1}textbf{V}^T)textbf{V}Sigmatextbf{U}^T$
$textbf{V}Sigma^{-1}textbf{U}^T$
$= (textbf{A}^T){^{-1}}$
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
edited Dec 6 '18 at 6:46
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
answered Dec 6 '18 at 5:09
W. MacTurkW. MacTurk
105
105
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
|
show 6 more comments
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Your reduction does not make sense unless $A^TA$ is invertible.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:23
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
Sorry, it was shown in a previous part that $(mathbf A^Tmathbf A+εmathbf I)$ is invertible, and my "reduction" (though not strictly true) hold in the limit -- since the $- epsilon textbf{I}$ part vanishes in the limit, $(mathbf A^Tmathbf A)$ is infinitesimally close to $(mathbf A^Tmathbf A+εmathbf I)$ while still being invertible. I just leave off the epsilons since, in the limit, they make no contribution.
$endgroup$
– W. MacTurk
Dec 6 '18 at 6:43
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
It sounds like you removed $varepsilon$ after some steps because removing it does not matter. I want to say, it does matter. We even do not know $A$ is a square matrix so the inverse could exist.
$endgroup$
– Hanul Jeon
Dec 7 '18 at 5:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
$textbf{A}^Ttextbf{A}$ is always a square matrix, is it not?
$endgroup$
– W. MacTurk
Dec 8 '18 at 6:05
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
$begingroup$
It is right, but you finally concludes the result of your reduction is $(A^T)^{-1}$.
$endgroup$
– Hanul Jeon
Dec 8 '18 at 6:07
|
show 6 more comments
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$begingroup$
Are you familiar with Singular Value Decomposition?
$endgroup$
– JimmyK4542
Dec 6 '18 at 2:46
$begingroup$
You can compute the inverse explicitly from the SVD, which is exactly what showed up in the answer to your previous question. The largest singular value is then 1/eps and this blows up
$endgroup$
– whpowell96
Dec 6 '18 at 3:39
$begingroup$
I know you can compute the inverse explicitly, but the problem is in an abstract form. I'm asked to find the limit of the expression, which reduces to $textbf{A}^Htextbf{A})^{-1}textbf{A}^T$. I am not sure where to go from here - I'm assuming this expression reduces further to something obvious but am not sure what that is.
$endgroup$
– W. MacTurk
Dec 6 '18 at 4:42
2
$begingroup$
$A^TA$ is not necessarily invertible. The expression in question is a well-known limit expression for the Moore-Penrose inverse of $A$.
$endgroup$
– user1551
Dec 6 '18 at 5:12