Solving for a matrix using least squares
0
$begingroup$
I am trying to understand equation 26 given equation 25.
I know that generally, if we have an overdetermined system of linear equations of the form
$Ax = b$
the least squares solution is
$hat{x} = (A^TA)^{-1}A^Tb$
Applying it to the above equation of
$D [R|t] = C$
where $[R|t]$ is the unknown matrix, we get
$hat{[R|t]} = (D^TD)^{-1}D^TC$
but that's not exactly what's shown in equation 26. How did (26) come from (25) if they are "solving linearly"?
K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
Thanks!
linear-algebra
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
$endgroup$
add a comment |
0
$begingroup$
I am trying to understand equation 26 given equation 25.
I know that generally, if we have an overdetermined system of linear equations of the form
$Ax = b$
the least squares solution is
$hat{x} = (A^TA)^{-1}A^Tb$
Applying it to the above equation of
$D [R|t] = C$
where $[R|t]$ is the unknown matrix, we get
$hat{[R|t]} = (D^TD)^{-1}D^TC$
but that's not exactly what's shown in equation 26. How did (26) come from (25) if they are "solving linearly"?
K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
Thanks!
linear-algebra
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
$endgroup$
$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15
add a comment |
0
0
0
$begingroup$
I am trying to understand equation 26 given equation 25.
I know that generally, if we have an overdetermined system of linear equations of the form
$Ax = b$
the least squares solution is
$hat{x} = (A^TA)^{-1}A^Tb$
Applying it to the above equation of
$D [R|t] = C$
where $[R|t]$ is the unknown matrix, we get
$hat{[R|t]} = (D^TD)^{-1}D^TC$
but that's not exactly what's shown in equation 26. How did (26) come from (25) if they are "solving linearly"?
K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
Thanks!
linear-algebra
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
$endgroup$
I am trying to understand equation 26 given equation 25.
I know that generally, if we have an overdetermined system of linear equations of the form
$Ax = b$
the least squares solution is
$hat{x} = (A^TA)^{-1}A^Tb$
Applying it to the above equation of
$D [R|t] = C$
where $[R|t]$ is the unknown matrix, we get
$hat{[R|t]} = (D^TD)^{-1}D^TC$
but that's not exactly what's shown in equation 26. How did (26) come from (25) if they are "solving linearly"?
K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
Thanks!
linear-algebra
linear-algebra
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
edited Dec 20 '18 at 6:37
Carpetfizz
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
asked Dec 20 '18 at 6:17
CarpetfizzCarpetfizz
486313
486313
$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15
add a comment |
$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15
$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15
add a comment |
1 Answer
1
active
oldest
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$begingroup$
In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $hat{[R|t]}= CD^T*(DD^T)^{-1}$
answered Dec 20 '18 at 8:29
user617446user617446
4443
$endgroup$
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $hat{[R|t]}= CD^T*(DD^T)^{-1}$
answered Dec 20 '18 at 8:29
user617446user617446
4443
$endgroup$
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
add a comment |
1
$begingroup$
In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $hat{[R|t]}= CD^T*(DD^T)^{-1}$
answered Dec 20 '18 at 8:29
user617446user617446
4443
$endgroup$
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
add a comment |
1
1
1
$begingroup$
In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $hat{[R|t]}= CD^T*(DD^T)^{-1}$
answered Dec 20 '18 at 8:29
user617446user617446
4443
$endgroup$
In fact, equation 25 is $ [R|t] D = C$ and not $D [R|t] = C$ , and indeed $ [R|t] DD^T = CD^T$ , yielding $hat{[R|t]}= CD^T*(DD^T)^{-1}$
answered Dec 20 '18 at 8:29
user617446user617446
4443
answered Dec 20 '18 at 8:29
user617446user617446
4443
answered Dec 20 '18 at 8:29
user617446user617446
4443
answered Dec 20 '18 at 8:29
user617446user617446
4443
4443
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
add a comment |
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
How did you get the second and third expressions? I know they match the text in the OP, but are you actually applying the least squares equation to get to it?
$endgroup$
– Carpetfizz
Dec 20 '18 at 8:31
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
I start from $[R|t]D=C$ and just right multiply by $D^T$ . The least squares is valid in exactly the same way, just consider $x^T A=b^T$ in your original definition.
$endgroup$
– user617446
Dec 20 '18 at 9:53
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Thank you, do you mind pointing me to a reference that talks about how to solve $x^TA = b^T$? Does the proof work the same way as the proof for solving $Ax = b$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 1:32
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
$begingroup$
Also, did you mean "consider $x^TA^T = b^T$ ?
$endgroup$
– Carpetfizz
Dec 21 '18 at 2:11
1
1
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
$begingroup$
The proof works the same way, since $x^T A-b^T = (A^T x-b)^T$ and the object to minimize, the norm squared, is the same.
$endgroup$
– user617446
Dec 23 '18 at 4:16
add a comment |
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$begingroup$
Can you define what $K$, $R$, and $t$ are?
$endgroup$
– Wolfy
Dec 20 '18 at 6:34
$begingroup$
Yes, sorry. K is a 3x3 “camera matrix”, R is a 3x3 rotation matrix, and t is a 3x1 translation vector
$endgroup$
– Carpetfizz
Dec 20 '18 at 6:35
$begingroup$
Okay, please edit your question, so other people know as well :).
$endgroup$
– Wolfy
Dec 20 '18 at 6:36
$begingroup$
These are two forms of the pseudo inverse matrix. We select one or the other depending on the dimensions of $D$, when $D$ is not square. When $D$ is square and full rank, no difference. An easy demonstration if $D$ is square: consider SVD ...
$endgroup$
– Damien
Dec 20 '18 at 9:12
$begingroup$
A precision: if $D$ is not square, one of the two forms is invalid ...
$endgroup$
– Damien
Dec 20 '18 at 9:15