Do trace properties of matrix hold for $(1,1)-$ tensors

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If $h, k$ be $(1,1)-$tensors on a Riemannian manifold $(M,g)$ then (similar to matrices) does the following property hold?
$$operatornametr(hcirc k)=operatornametr(kcirc h).$$

More generally if $s$ be another $(1,1)-$tensor then does the following property hold?
$$operatornametr (hcirc k circ s)=operatornametr(scirc hcirc k).$$

linear-algebra linear-transformations tensors

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edited Aug 6 at 15:05

Bernard

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asked Aug 6 at 14:58

C.F.G

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up vote
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If $h, k$ be $(1,1)-$tensors on a Riemannian manifold $(M,g)$ then (similar to matrices) does the following property hold?
$$operatornametr(hcirc k)=operatornametr(kcirc h).$$

More generally if $s$ be another $(1,1)-$tensor then does the following property hold?
$$operatornametr (hcirc k circ s)=operatornametr(scirc hcirc k).$$

linear-algebra linear-transformations tensors

share|cite|improve this question

edited Aug 6 at 15:05

Bernard

110k635103

asked Aug 6 at 14:58

C.F.G

1,3111521

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If $h, k$ be $(1,1)-$tensors on a Riemannian manifold $(M,g)$ then (similar to matrices) does the following property hold?
$$operatornametr(hcirc k)=operatornametr(kcirc h).$$

More generally if $s$ be another $(1,1)-$tensor then does the following property hold?
$$operatornametr (hcirc k circ s)=operatornametr(scirc hcirc k).$$

linear-algebra linear-transformations tensors

share|cite|improve this question

edited Aug 6 at 15:05

Bernard

110k635103

asked Aug 6 at 14:58

C.F.G

1,3111521

If $h, k$ be $(1,1)-$tensors on a Riemannian manifold $(M,g)$ then (similar to matrices) does the following property hold?
$$operatornametr(hcirc k)=operatornametr(kcirc h).$$

More generally if $s$ be another $(1,1)-$tensor then does the following property hold?
$$operatornametr (hcirc k circ s)=operatornametr(scirc hcirc k).$$

linear-algebra linear-transformations tensors

share|cite|improve this question

edited Aug 6 at 15:05

Bernard

110k635103

asked Aug 6 at 14:58

C.F.G

1,3111521

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 15:05

Bernard

110k635103

edited Aug 6 at 15:05

Bernard

110k635103

edited Aug 6 at 15:05

Bernard

110k635103

110k635103

asked Aug 6 at 14:58

C.F.G

1,3111521

asked Aug 6 at 14:58

C.F.G

1,3111521

asked Aug 6 at 14:58

C.F.G

1,3111521

1,3111521

add a comment | 

add a comment | 

1 Answer
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accepted

Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

add a comment | 

up vote
3
down vote



accepted

Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

answered Aug 6 at 15:08

Alfred Yerger

9,7952044

9,7952044

add a comment | 

add a comment | 

 

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