There are no $3$ linearly independent lightlike vectors such that $u+v+w = 0$.
0
$begingroup$
Consider the Lorentz-Minkowski space $E^n_1$, also known as $mathbb{L}^n$. I want to prove that there are not lightlike linearly independent vectors $u, v, w in E^n_1$ such that $u + v + w = 0$. How to do it? I'm still unfamiliar with the intuition behind such space.
linear-algebra semi-riemannian-geometry
asked Dec 8 '18 at 1:31
user71487user71487
948
$endgroup$
|
show 1 more comment
0
$begingroup$
Consider the Lorentz-Minkowski space $E^n_1$, also known as $mathbb{L}^n$. I want to prove that there are not lightlike linearly independent vectors $u, v, w in E^n_1$ such that $u + v + w = 0$. How to do it? I'm still unfamiliar with the intuition behind such space.
linear-algebra semi-riemannian-geometry
asked Dec 8 '18 at 1:31
user71487user71487
948
$endgroup$
$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44
|
show 1 more comment
0
0
0
1
$begingroup$
Consider the Lorentz-Minkowski space $E^n_1$, also known as $mathbb{L}^n$. I want to prove that there are not lightlike linearly independent vectors $u, v, w in E^n_1$ such that $u + v + w = 0$. How to do it? I'm still unfamiliar with the intuition behind such space.
linear-algebra semi-riemannian-geometry
asked Dec 8 '18 at 1:31
user71487user71487
948
$endgroup$
Consider the Lorentz-Minkowski space $E^n_1$, also known as $mathbb{L}^n$. I want to prove that there are not lightlike linearly independent vectors $u, v, w in E^n_1$ such that $u + v + w = 0$. How to do it? I'm still unfamiliar with the intuition behind such space.
linear-algebra semi-riemannian-geometry
linear-algebra semi-riemannian-geometry
asked Dec 8 '18 at 1:31
user71487user71487
948
asked Dec 8 '18 at 1:31
user71487user71487
948
edited Dec 8 '18 at 1:45
user71487
asked Dec 8 '18 at 1:31
user71487user71487
948
asked Dec 8 '18 at 1:31
user71487user71487
948
asked Dec 8 '18 at 1:31
user71487user71487
948
948
$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44
|
show 1 more comment
$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44
$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
Hint Suppose there were. Expand $$0 = [{bf u} + {bf v} + {bf w}] cdot [{bf u} - ({bf v} + {bf w})]$$ to conclude that ${bf v} cdot {bf w} = 0$.
Additional hint What is the matrix representation of the bilinear form $cdot$ with respect to the basis $({bf u}, {bf v}, {bf w})$?
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
$endgroup$
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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1
$begingroup$
Hint Suppose there were. Expand $$0 = [{bf u} + {bf v} + {bf w}] cdot [{bf u} - ({bf v} + {bf w})]$$ to conclude that ${bf v} cdot {bf w} = 0$.
Additional hint What is the matrix representation of the bilinear form $cdot$ with respect to the basis $({bf u}, {bf v}, {bf w})$?
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
$endgroup$
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
|
show 2 more comments
1
$begingroup$
Hint Suppose there were. Expand $$0 = [{bf u} + {bf v} + {bf w}] cdot [{bf u} - ({bf v} + {bf w})]$$ to conclude that ${bf v} cdot {bf w} = 0$.
Additional hint What is the matrix representation of the bilinear form $cdot$ with respect to the basis $({bf u}, {bf v}, {bf w})$?
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
$endgroup$
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
|
show 2 more comments
1
1
1
$begingroup$
Hint Suppose there were. Expand $$0 = [{bf u} + {bf v} + {bf w}] cdot [{bf u} - ({bf v} + {bf w})]$$ to conclude that ${bf v} cdot {bf w} = 0$.
Additional hint What is the matrix representation of the bilinear form $cdot$ with respect to the basis $({bf u}, {bf v}, {bf w})$?
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
$endgroup$
Hint Suppose there were. Expand $$0 = [{bf u} + {bf v} + {bf w}] cdot [{bf u} - ({bf v} + {bf w})]$$ to conclude that ${bf v} cdot {bf w} = 0$.
Additional hint What is the matrix representation of the bilinear form $cdot$ with respect to the basis $({bf u}, {bf v}, {bf w})$?
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
answered Dec 8 '18 at 1:43
TravisTravis
59.9k767146
59.9k767146
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
|
show 2 more comments
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({bf u}, {bf v}, {bf w})$ is not a basis of $mathbb L^n$ for $n neq 3$.
$endgroup$
– Travis
Dec 8 '18 at 1:48
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
Shouldn't I conclude ${bf u} cdot {bf w} = 0$ instead?
$endgroup$
– user71487
Dec 8 '18 at 1:58
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
$begingroup$
I don't see how---after all, the expression is symmetric in ${bf v}$ and ${bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:02
1
1
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
$begingroup$
Expanding gives $0 = {bf u} cdot {bf u} - ({bf v} + {bf w}) cdot ({bf v} + {bf w}) = -({bf v} cdot {bf v} + 2 {bf v} cdot {bf w} + {bf w} cdot {bf w}) = -2 {bf v} cdot {bf w}$.
$endgroup$
– Travis
Dec 8 '18 at 2:58
1
1
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
$begingroup$
Not directly anyway. By symmetry of notation, we also have ${bf u} cdot {bf v} = {bf w} cdot {bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $cdot$ is the zero bilinear form, a contradiction.
$endgroup$
– Travis
Dec 8 '18 at 3:00
|
show 2 more comments
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$begingroup$
Does this space have 2 spacelike and one timelike variables? or what if not?
$endgroup$
– coffeemath
Dec 8 '18 at 1:38
$begingroup$
@coffeemath Those are all hypotheses I have.
$endgroup$
– user71487
Dec 8 '18 at 1:41
$begingroup$
I don't see how this can be true... can't we take one of the vectors from the future light cone, say $vec{u}$, and the other two to be $-frac{1}{2}vec{u}$? Don't we need to say that they are non-coplanar?
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:41
$begingroup$
@RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent.
$endgroup$
– Travis
Dec 8 '18 at 1:44
$begingroup$
@Travis I should learn to read the titles more carefully!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 1:44