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Volume of an $n$-dimensional ellipsoid
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up vote
2
down vote
favorite
Consider the convex body $S_r subset mathbbR^n$ defined as follows
$$S_r := mathbbx in mathbbR^n : $$
for some invertible $n times n$ matrix $A$ and some real constant $r > 0$. Here, $|cdot |$ denotes the Euclidean norm. How is the volume of $S_r$ calculated?
linear-algebra geometry volume
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
asked Jul 17 at 21:09
Chris
276112
add a comment |Â
up vote
2
down vote
favorite
Consider the convex body $S_r subset mathbbR^n$ defined as follows
$$S_r := mathbbx in mathbbR^n : $$
for some invertible $n times n$ matrix $A$ and some real constant $r > 0$. Here, $|cdot |$ denotes the Euclidean norm. How is the volume of $S_r$ calculated?
linear-algebra geometry volume
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
asked Jul 17 at 21:09
Chris
276112
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the convex body $S_r subset mathbbR^n$ defined as follows
$$S_r := mathbbx in mathbbR^n : $$
for some invertible $n times n$ matrix $A$ and some real constant $r > 0$. Here, $|cdot |$ denotes the Euclidean norm. How is the volume of $S_r$ calculated?
linear-algebra geometry volume
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
asked Jul 17 at 21:09
Chris
276112
Consider the convex body $S_r subset mathbbR^n$ defined as follows
$$S_r := mathbbx in mathbbR^n : $$
for some invertible $n times n$ matrix $A$ and some real constant $r > 0$. Here, $|cdot |$ denotes the Euclidean norm. How is the volume of $S_r$ calculated?
linear-algebra geometry volume
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
asked Jul 17 at 21:09
Chris
276112
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
edited Jul 18 at 5:59
Rodrigo de Azevedo
12.5k41751
12.5k41751
asked Jul 17 at 21:09
Chris
276112
asked Jul 17 at 21:09
Chris
276112
asked Jul 17 at 21:09
Chris
276112
276112
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x in S_r$ iff $|x^T A^TA x| le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $Lambda$
such that $A^TA = U Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $ sum_k lambda_k x_k^2 le r^2 = sum_k ( sqrtlambda_k x_k)^2 le r^2 = $.
It is straightforward to see that the volume of the last set is
$r^n over sqrtlambda_1 cdots lambda_n = r^n over $ times the
volume of the unit ball $overlineB(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $mathbbR^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $lambda$), it
is straightforward to show that $m(L(A)) = |lambda| m(A)$. Since the map
$x mapsto (r x_1 over sqrtlambda_1,...,r x_n over sqrtlambda_n)$
can be written as the composition of $n$ such operations, we see that
begineqnarray
m( ) &=& |r over sqrtlambda_1|m((x_1,...,r x_n over sqrtlambda_n) ) \
&vdots& \
&=& r^n over sqrtlambda_1 cdots lambda_n m( x)
endeqnarray
And more:
To see why $m(L(A)) = |lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = inf sum_k m(R_k) $.
It is not difficult to see that $m(L(R)) = |lambda| m(R)$ for a rectangle.
If $cal R_A= R_k_k $, then it is straightforward to see that
$cal R_L(A) = R_k_k in cal R_A $.
Hence $m(L(A)) = inf R_k_k text is a cover of A text by rectangles = |lambda| m(A)$.
answered Jul 17 at 22:40
copper.hat
122k557156
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
 |Â
show 11 more comments
up vote
2
down vote
Use the substitution rule (where $phi : Btophi(B)$ is a diffeomorphism)
$$
int_phi(B)f(x),dx = int_Bf(phi(y))|detphi'(y)|,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $phi(y) = A^-1y$.
answered Jul 17 at 21:16
amsmath
1,613114
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
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
Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x in S_r$ iff $|x^T A^TA x| le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $Lambda$
such that $A^TA = U Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $ sum_k lambda_k x_k^2 le r^2 = sum_k ( sqrtlambda_k x_k)^2 le r^2 = $.
It is straightforward to see that the volume of the last set is
$r^n over sqrtlambda_1 cdots lambda_n = r^n over $ times the
volume of the unit ball $overlineB(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $mathbbR^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $lambda$), it
is straightforward to show that $m(L(A)) = |lambda| m(A)$. Since the map
$x mapsto (r x_1 over sqrtlambda_1,...,r x_n over sqrtlambda_n)$
can be written as the composition of $n$ such operations, we see that
begineqnarray
m( ) &=& |r over sqrtlambda_1|m((x_1,...,r x_n over sqrtlambda_n) ) \
&vdots& \
&=& r^n over sqrtlambda_1 cdots lambda_n m( x)
endeqnarray
And more:
To see why $m(L(A)) = |lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = inf sum_k m(R_k) $.
It is not difficult to see that $m(L(R)) = |lambda| m(R)$ for a rectangle.
If $cal R_A= R_k_k $, then it is straightforward to see that
$cal R_L(A) = R_k_k in cal R_A $.
Hence $m(L(A)) = inf R_k_k text is a cover of A text by rectangles = |lambda| m(A)$.
answered Jul 17 at 22:40
copper.hat
122k557156
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
 |Â
show 11 more comments
up vote
1
down vote
accepted
Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x in S_r$ iff $|x^T A^TA x| le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $Lambda$
such that $A^TA = U Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $ sum_k lambda_k x_k^2 le r^2 = sum_k ( sqrtlambda_k x_k)^2 le r^2 = $.
It is straightforward to see that the volume of the last set is
$r^n over sqrtlambda_1 cdots lambda_n = r^n over $ times the
volume of the unit ball $overlineB(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $mathbbR^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $lambda$), it
is straightforward to show that $m(L(A)) = |lambda| m(A)$. Since the map
$x mapsto (r x_1 over sqrtlambda_1,...,r x_n over sqrtlambda_n)$
can be written as the composition of $n$ such operations, we see that
begineqnarray
m( ) &=& |r over sqrtlambda_1|m((x_1,...,r x_n over sqrtlambda_n) ) \
&vdots& \
&=& r^n over sqrtlambda_1 cdots lambda_n m( x)
endeqnarray
And more:
To see why $m(L(A)) = |lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = inf sum_k m(R_k) $.
It is not difficult to see that $m(L(R)) = |lambda| m(R)$ for a rectangle.
If $cal R_A= R_k_k $, then it is straightforward to see that
$cal R_L(A) = R_k_k in cal R_A $.
Hence $m(L(A)) = inf R_k_k text is a cover of A text by rectangles = |lambda| m(A)$.
answered Jul 17 at 22:40
copper.hat
122k557156
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
 |Â
show 11 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x in S_r$ iff $|x^T A^TA x| le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $Lambda$
such that $A^TA = U Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $ sum_k lambda_k x_k^2 le r^2 = sum_k ( sqrtlambda_k x_k)^2 le r^2 = $.
It is straightforward to see that the volume of the last set is
$r^n over sqrtlambda_1 cdots lambda_n = r^n over $ times the
volume of the unit ball $overlineB(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $mathbbR^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $lambda$), it
is straightforward to show that $m(L(A)) = |lambda| m(A)$. Since the map
$x mapsto (r x_1 over sqrtlambda_1,...,r x_n over sqrtlambda_n)$
can be written as the composition of $n$ such operations, we see that
begineqnarray
m( ) &=& |r over sqrtlambda_1|m((x_1,...,r x_n over sqrtlambda_n) ) \
&vdots& \
&=& r^n over sqrtlambda_1 cdots lambda_n m( x)
endeqnarray
And more:
To see why $m(L(A)) = |lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = inf sum_k m(R_k) $.
It is not difficult to see that $m(L(R)) = |lambda| m(R)$ for a rectangle.
If $cal R_A= R_k_k $, then it is straightforward to see that
$cal R_L(A) = R_k_k in cal R_A $.
Hence $m(L(A)) = inf R_k_k text is a cover of A text by rectangles = |lambda| m(A)$.
answered Jul 17 at 22:40
copper.hat
122k557156
Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x in S_r$ iff $|x^T A^TA x| le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $Lambda$
such that $A^TA = U Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $ sum_k lambda_k x_k^2 le r^2 = sum_k ( sqrtlambda_k x_k)^2 le r^2 = $.
It is straightforward to see that the volume of the last set is
$r^n over sqrtlambda_1 cdots lambda_n = r^n over $ times the
volume of the unit ball $overlineB(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $mathbbR^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $lambda$), it
is straightforward to show that $m(L(A)) = |lambda| m(A)$. Since the map
$x mapsto (r x_1 over sqrtlambda_1,...,r x_n over sqrtlambda_n)$
can be written as the composition of $n$ such operations, we see that
begineqnarray
m( ) &=& |r over sqrtlambda_1|m((x_1,...,r x_n over sqrtlambda_n) ) \
&vdots& \
&=& r^n over sqrtlambda_1 cdots lambda_n m( x)
endeqnarray
And more:
To see why $m(L(A)) = |lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = inf sum_k m(R_k) $.
It is not difficult to see that $m(L(R)) = |lambda| m(R)$ for a rectangle.
If $cal R_A= R_k_k $, then it is straightforward to see that
$cal R_L(A) = R_k_k in cal R_A $.
Hence $m(L(A)) = inf R_k_k text is a cover of A text by rectangles = |lambda| m(A)$.
answered Jul 17 at 22:40
copper.hat
122k557156
edited Jul 18 at 23:17
answered Jul 17 at 22:40
copper.hat
122k557156
answered Jul 17 at 22:40
copper.hat
122k557156
answered Jul 17 at 22:40
copper.hat
122k557156
122k557156
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
 |Â
show 11 more comments
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
I don't see that finding out the volume of the last set is straightforward. How do you do that without integrating?
– amsmath
Jul 18 at 1:51
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: Do you mean the one with $|x| le 1$?
– copper.hat
Jul 18 at 3:06
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
@amsmath: I'm not exactly sure what you are asking, at some level computing the measure of a set is essentially integrating. In any event, I elaborated the computation of the last set, hopefully that is what you were asking?
– copper.hat
Jul 18 at 3:18
1
1
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
Sorry for the late reply - thanks for the answer @copper.hat!
– Chris
Jul 18 at 16:33
1
1
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
@copper.hat Thank you very much. This explains a lot.
– amsmath
Jul 18 at 23:24
 |Â
show 11 more comments
up vote
2
down vote
Use the substitution rule (where $phi : Btophi(B)$ is a diffeomorphism)
$$
int_phi(B)f(x),dx = int_Bf(phi(y))|detphi'(y)|,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $phi(y) = A^-1y$.
answered Jul 17 at 21:16
amsmath
1,613114
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
add a comment |Â
up vote
2
down vote
Use the substitution rule (where $phi : Btophi(B)$ is a diffeomorphism)
$$
int_phi(B)f(x),dx = int_Bf(phi(y))|detphi'(y)|,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $phi(y) = A^-1y$.
answered Jul 17 at 21:16
amsmath
1,613114
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Use the substitution rule (where $phi : Btophi(B)$ is a diffeomorphism)
$$
int_phi(B)f(x),dx = int_Bf(phi(y))|detphi'(y)|,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $phi(y) = A^-1y$.
answered Jul 17 at 21:16
amsmath
1,613114
Use the substitution rule (where $phi : Btophi(B)$ is a diffeomorphism)
$$
int_phi(B)f(x),dx = int_Bf(phi(y))|detphi'(y)|,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $phi(y) = A^-1y$.
answered Jul 17 at 21:16
amsmath
1,613114
answered Jul 17 at 21:16
amsmath
1,613114
answered Jul 17 at 21:16
amsmath
1,613114
answered Jul 17 at 21:16
amsmath
1,613114
1,613114
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
add a comment |Â
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
Thanks for the answer @amsmath!
– Chris
Jul 18 at 16:35
add a comment |Â
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