Mixing
Invertible Matrices isomorphisms vs projections
Clash Royale CLAN TAG#URR8PPP
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I'm self learning linear-algebra.
I'm having trouble understanding why if $A$ is a matrix where $F(x) = Ax$ is an orthogonal projection, it cannot be necessarily invertible.
Why is that the case, but when $A$ is a matrix such that $F(x) = Ax$ is an isomorphism, $A$ is necessarily invertible?
linear-algebra matrices linear-transformations
edited Jul 22 at 9:52
Davide Morgante
1,800220
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
 |Â
show 2 more comments
up vote
1
down vote
favorite
I'm self learning linear-algebra.
I'm having trouble understanding why if $A$ is a matrix where $F(x) = Ax$ is an orthogonal projection, it cannot be necessarily invertible.
Why is that the case, but when $A$ is a matrix such that $F(x) = Ax$ is an isomorphism, $A$ is necessarily invertible?
linear-algebra matrices linear-transformations
edited Jul 22 at 9:52
Davide Morgante
1,800220
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
1
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm self learning linear-algebra.
I'm having trouble understanding why if $A$ is a matrix where $F(x) = Ax$ is an orthogonal projection, it cannot be necessarily invertible.
Why is that the case, but when $A$ is a matrix such that $F(x) = Ax$ is an isomorphism, $A$ is necessarily invertible?
linear-algebra matrices linear-transformations
edited Jul 22 at 9:52
Davide Morgante
1,800220
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
I'm self learning linear-algebra.
I'm having trouble understanding why if $A$ is a matrix where $F(x) = Ax$ is an orthogonal projection, it cannot be necessarily invertible.
Why is that the case, but when $A$ is a matrix such that $F(x) = Ax$ is an isomorphism, $A$ is necessarily invertible?
linear-algebra matrices linear-transformations
edited Jul 22 at 9:52
Davide Morgante
1,800220
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
edited Jul 22 at 9:52
Davide Morgante
1,800220
edited Jul 22 at 9:52
Davide Morgante
1,800220
edited Jul 22 at 9:52
Davide Morgante
1,800220
1,800220
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
asked Jul 22 at 9:45
PERTURBATIONFLOW
396
396
1
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22
 |Â
show 2 more comments
1
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22
1
1
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Imagine you want to project all $mathbbR^3$ onto a plane $pi$: necessarily some points from $mathbbR^3$ have to go to same points in $pi$ (simply speaking there are too many points in $mathbbR^3$, you have to squeeze $mathbbR^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $pi$ from it's respective point, where they came from, in $mathbbR^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $pi$ it has to get you again to $pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^-1$ that gives you the inverse map
answered Jul 22 at 9:56
Davide Morgante
1,800220
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
add a comment |Â
up vote
1
down vote
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^-1 $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
answered Jul 22 at 9:55
Martin Argerami
116k1071164
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
Imagine you want to project all $mathbbR^3$ onto a plane $pi$: necessarily some points from $mathbbR^3$ have to go to same points in $pi$ (simply speaking there are too many points in $mathbbR^3$, you have to squeeze $mathbbR^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $pi$ from it's respective point, where they came from, in $mathbbR^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $pi$ it has to get you again to $pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^-1$ that gives you the inverse map
answered Jul 22 at 9:56
Davide Morgante
1,800220
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
add a comment |Â
up vote
1
down vote
accepted
Imagine you want to project all $mathbbR^3$ onto a plane $pi$: necessarily some points from $mathbbR^3$ have to go to same points in $pi$ (simply speaking there are too many points in $mathbbR^3$, you have to squeeze $mathbbR^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $pi$ from it's respective point, where they came from, in $mathbbR^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $pi$ it has to get you again to $pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^-1$ that gives you the inverse map
answered Jul 22 at 9:56
Davide Morgante
1,800220
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Imagine you want to project all $mathbbR^3$ onto a plane $pi$: necessarily some points from $mathbbR^3$ have to go to same points in $pi$ (simply speaking there are too many points in $mathbbR^3$, you have to squeeze $mathbbR^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $pi$ from it's respective point, where they came from, in $mathbbR^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $pi$ it has to get you again to $pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^-1$ that gives you the inverse map
answered Jul 22 at 9:56
Davide Morgante
1,800220
Imagine you want to project all $mathbbR^3$ onto a plane $pi$: necessarily some points from $mathbbR^3$ have to go to same points in $pi$ (simply speaking there are too many points in $mathbbR^3$, you have to squeeze $mathbbR^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $pi$ from it's respective point, where they came from, in $mathbbR^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $pi$ it has to get you again to $pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^-1$ that gives you the inverse map
answered Jul 22 at 9:56
Davide Morgante
1,800220
edited Jul 22 at 10:03
answered Jul 22 at 9:56
Davide Morgante
1,800220
answered Jul 22 at 9:56
Davide Morgante
1,800220
answered Jul 22 at 9:56
Davide Morgante
1,800220
1,800220
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
add a comment |Â
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
as a follow up question, how does one disprove that any symmetric matrix is invertible
– PERTURBATIONFLOW
Jul 22 at 9:58
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
I can give you a simple example $$left[beginmatrix1&1\1&1endmatrixright]$$ is clearly symmetric but not invertible, in fact it's determinant is zero
– Davide Morgante
Jul 22 at 10:00
add a comment |Â
up vote
1
down vote
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^-1 $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
answered Jul 22 at 9:55
Martin Argerami
116k1071164
add a comment |Â
up vote
1
down vote
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^-1 $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
answered Jul 22 at 9:55
Martin Argerami
116k1071164
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^-1 $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
answered Jul 22 at 9:55
Martin Argerami
116k1071164
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^-1 $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
answered Jul 22 at 9:55
Martin Argerami
116k1071164
answered Jul 22 at 9:55
Martin Argerami
116k1071164
answered Jul 22 at 9:55
Martin Argerami
116k1071164
answered Jul 22 at 9:55
Martin Argerami
116k1071164
116k1071164
add a comment |Â
add a comment |Â
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1
A matrix like $A=pmatrix1&0&0\0&1&0\0&0&0$ which projects $Bbb R^3$ orthogonally onto a plane, perhaps?
– Lord Shark the Unknown
Jul 22 at 9:46
@LordSharktheUnknown can you please clarify your statement?
– PERTURBATIONFLOW
Jul 22 at 9:53
Isomorphisms are invertible by definition
– A. Goodier
Jul 22 at 9:56
Just as an aside, here oh MSE you should mark an answer as accepted if you feel that it gave you what you were looking for. This not only gives credit to the user that wrote the answer, but it marks your post as closed to every other MSE user
– Davide Morgante
Jul 22 at 10:15
@DavideMorgante. As you can tell, I'm a quite new user. How do I do that? Thank you for all your assistance
– PERTURBATIONFLOW
Jul 22 at 10:22