Is it really true that U^-1 = U^T when U is an orthogonal square matrix?
I read my textbook and it says $ U^{-1} = U^{T}$ when U is an orthogonal square matrix.
I pick this matrix
but it doesn't seems the theorem holds.
linear-algebra matrices
asked Nov 12 '16 at 18:31
TmmTmm
698717
|
show 1 more comment
I read my textbook and it says $ U^{-1} = U^{T}$ when U is an orthogonal square matrix.
I pick this matrix
but it doesn't seems the theorem holds.
linear-algebra matrices
asked Nov 12 '16 at 18:31
TmmTmm
698717
2
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
3
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
1
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20
|
show 1 more comment
I read my textbook and it says $ U^{-1} = U^{T}$ when U is an orthogonal square matrix.
I pick this matrix
but it doesn't seems the theorem holds.
linear-algebra matrices
asked Nov 12 '16 at 18:31
TmmTmm
698717
I read my textbook and it says $ U^{-1} = U^{T}$ when U is an orthogonal square matrix.
I pick this matrix
but it doesn't seems the theorem holds.
linear-algebra matrices
linear-algebra matrices
asked Nov 12 '16 at 18:31
TmmTmm
698717
asked Nov 12 '16 at 18:31
TmmTmm
698717
asked Nov 12 '16 at 18:31
TmmTmm
698717
asked Nov 12 '16 at 18:31
TmmTmm
698717
asked Nov 12 '16 at 18:31
TmmTmm
698717
698717
2
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
3
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
1
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20
|
show 1 more comment
2
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
3
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
1
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20
2
2
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
3
3
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
1
1
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20
|
show 1 more comment
2 Answers
2
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Let me suppose your definition of an orthogonal matrix $A$ is that $A^TA=I$ must hold. It is generally true for square matrices $AB$ (over a field, or even over commutative ring) that $BA=I$ implies $AB=I$. So we have $AA^T=I$ as well, and $A^T$ satisfies the definition $A^{-1}$. (And inverses are unique if they exist.)
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
The problem is that your given matrix is not orthogonal. An orthogonal matrix has orthogonal column vectors, meaning that their inner product should be zero. This is not the case for your matrix. If you take the inner product of the first and third row from your matrix you get 2, not 0.
answered 2 days ago
GustafGustaf
12
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2 Answers
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active
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2 Answers
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active
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votes
Let me suppose your definition of an orthogonal matrix $A$ is that $A^TA=I$ must hold. It is generally true for square matrices $AB$ (over a field, or even over commutative ring) that $BA=I$ implies $AB=I$. So we have $AA^T=I$ as well, and $A^T$ satisfies the definition $A^{-1}$. (And inverses are unique if they exist.)
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
Let me suppose your definition of an orthogonal matrix $A$ is that $A^TA=I$ must hold. It is generally true for square matrices $AB$ (over a field, or even over commutative ring) that $BA=I$ implies $AB=I$. So we have $AA^T=I$ as well, and $A^T$ satisfies the definition $A^{-1}$. (And inverses are unique if they exist.)
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
Let me suppose your definition of an orthogonal matrix $A$ is that $A^TA=I$ must hold. It is generally true for square matrices $AB$ (over a field, or even over commutative ring) that $BA=I$ implies $AB=I$. So we have $AA^T=I$ as well, and $A^T$ satisfies the definition $A^{-1}$. (And inverses are unique if they exist.)
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
Let me suppose your definition of an orthogonal matrix $A$ is that $A^TA=I$ must hold. It is generally true for square matrices $AB$ (over a field, or even over commutative ring) that $BA=I$ implies $AB=I$. So we have $AA^T=I$ as well, and $A^T$ satisfies the definition $A^{-1}$. (And inverses are unique if they exist.)
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Nov 13 '16 at 8:39
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
86.5k5106220
add a comment |
add a comment |
The problem is that your given matrix is not orthogonal. An orthogonal matrix has orthogonal column vectors, meaning that their inner product should be zero. This is not the case for your matrix. If you take the inner product of the first and third row from your matrix you get 2, not 0.
answered 2 days ago
GustafGustaf
12
New contributor
Gustaf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The problem is that your given matrix is not orthogonal. An orthogonal matrix has orthogonal column vectors, meaning that their inner product should be zero. This is not the case for your matrix. If you take the inner product of the first and third row from your matrix you get 2, not 0.
answered 2 days ago
GustafGustaf
12
New contributor
Gustaf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The problem is that your given matrix is not orthogonal. An orthogonal matrix has orthogonal column vectors, meaning that their inner product should be zero. This is not the case for your matrix. If you take the inner product of the first and third row from your matrix you get 2, not 0.
answered 2 days ago
GustafGustaf
12
New contributor
Gustaf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The problem is that your given matrix is not orthogonal. An orthogonal matrix has orthogonal column vectors, meaning that their inner product should be zero. This is not the case for your matrix. If you take the inner product of the first and third row from your matrix you get 2, not 0.
answered 2 days ago
GustafGustaf
12
New contributor
Gustaf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
GustafGustaf
12
New contributor
Gustaf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
GustafGustaf
12
answered 2 days ago
GustafGustaf
12
12
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New contributor
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2
What is your definition of orthogonal?
– lhf
Nov 12 '16 at 18:33
3
This is not an orthogonal matrix.
– Crostul
Nov 12 '16 at 18:34
@woogie I don't agree, computing the determinant is by far not the thing to do : it will take you time and even if you find $-1$ or $1$, it will prove nothing. You have to test whether the columns constitute an orthonormal basis, which is not the case : the norm of the first column is not equal to 1... immediate.
– Jean Marie
Nov 12 '16 at 18:52
@ JeanMarie @ Crostul
– Tmm
Nov 12 '16 at 20:06
1
@Tmm An orthogonal matrix is a matrix with orthonormal columns, which means the columns are orthogonal to each other and each have unit-length (length of 1). The columns of your matrix do not have length 1. By the way, "symmetric" means $A^T=A$ not $A^T=A^{-1}$. It is a common confusion to think an orthogonal matrix means "orthogonal columns," forgetting the length-1 part. If $A$ merely has orthogonal columns, then $A^TA$ is a diagonal matrix whose diagonal entries are the square-lengths of each of the columns.
– Kyle Miller
Nov 13 '16 at 9:20