Express 2-Tensor in Basis of $R^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$e_1otimes e_2 otimes e_3$ cannot be written as a sum of an alternating tensor and a symmetric tensorCondition for a tensor to be decomposableMatrix Representation of the Tensor Product of Linear MapsCalculate the tensor product of two vectorsTensor Product of Spaces has Basis of Tensor ProductsDoes the Euclidean dot product (via the tensor product) induce the trace?an element of a tensor product of modules not equal to any generator.Uniqueness of the basis for a tensor product of vector spaces.Convert from one tensor canonical form to anotherFinding non-singular transformation mapping one tensor to other in $(Bbb F_2)^otimes 3$
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Express 2-Tensor in Basis of $R^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$e_1otimes e_2 otimes e_3$ cannot be written as a sum of an alternating tensor and a symmetric tensorCondition for a tensor to be decomposableMatrix Representation of the Tensor Product of Linear MapsCalculate the tensor product of two vectorsTensor Product of Spaces has Basis of Tensor ProductsDoes the Euclidean dot product (via the tensor product) induce the trace?an element of a tensor product of modules not equal to any generator.Uniqueness of the basis for a tensor product of vector spaces.Convert from one tensor canonical form to anotherFinding non-singular transformation mapping one tensor to other in $(Bbb F_2)^otimes 3$
$begingroup$
I think I may just be confused by the question... would appreciate help!
I am given the function,
$f(ae_1 + be_2 , ce_1 + de_2) = ac + bd$
(it doesn't say, but I assume we have $e_1, e_2$ the standard basis for $R^2$)
First thing was to show that $f$ is bi-linear, which I have confirmed. Then I am to "expand the 2-tensor in the basis". Can someone explain what this means? Maybe I'm just not liking how it is stated. Meaning, write it as a sum $sum e_i otimes e_j$?
Secondly, show $e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = ac$.
This seems simple enough, as (my understanding is) $e_1$ "keeps" the 1st term.
$e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = e_1(ae_1 + be_2)cdot e_2(ce_1 + de_2) = a cdot c$
Third, express $(ae_1 + be_2) wedge (ce_1 + de_2)$ in terms of $ac(e_1 times e_2) + bd(e_2 times e_2)$. This should be tensor product, right? Not cross product, as it is stated.
My professor gives us sets of notes and problems to work on before going to class. He writes the questions himself, so sometimes the way they are stated doesn't seem clear to me.
tensor-products
asked Apr 9 at 3:08
HJoyHJoy
816
$endgroup$
add a comment |
$begingroup$
I think I may just be confused by the question... would appreciate help!
I am given the function,
$f(ae_1 + be_2 , ce_1 + de_2) = ac + bd$
(it doesn't say, but I assume we have $e_1, e_2$ the standard basis for $R^2$)
First thing was to show that $f$ is bi-linear, which I have confirmed. Then I am to "expand the 2-tensor in the basis". Can someone explain what this means? Maybe I'm just not liking how it is stated. Meaning, write it as a sum $sum e_i otimes e_j$?
Secondly, show $e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = ac$.
This seems simple enough, as (my understanding is) $e_1$ "keeps" the 1st term.
$e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = e_1(ae_1 + be_2)cdot e_2(ce_1 + de_2) = a cdot c$
Third, express $(ae_1 + be_2) wedge (ce_1 + de_2)$ in terms of $ac(e_1 times e_2) + bd(e_2 times e_2)$. This should be tensor product, right? Not cross product, as it is stated.
My professor gives us sets of notes and problems to work on before going to class. He writes the questions himself, so sometimes the way they are stated doesn't seem clear to me.
tensor-products
asked Apr 9 at 3:08
HJoyHJoy
816
$endgroup$
add a comment |
$begingroup$
I think I may just be confused by the question... would appreciate help!
I am given the function,
$f(ae_1 + be_2 , ce_1 + de_2) = ac + bd$
(it doesn't say, but I assume we have $e_1, e_2$ the standard basis for $R^2$)
First thing was to show that $f$ is bi-linear, which I have confirmed. Then I am to "expand the 2-tensor in the basis". Can someone explain what this means? Maybe I'm just not liking how it is stated. Meaning, write it as a sum $sum e_i otimes e_j$?
Secondly, show $e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = ac$.
This seems simple enough, as (my understanding is) $e_1$ "keeps" the 1st term.
$e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = e_1(ae_1 + be_2)cdot e_2(ce_1 + de_2) = a cdot c$
Third, express $(ae_1 + be_2) wedge (ce_1 + de_2)$ in terms of $ac(e_1 times e_2) + bd(e_2 times e_2)$. This should be tensor product, right? Not cross product, as it is stated.
My professor gives us sets of notes and problems to work on before going to class. He writes the questions himself, so sometimes the way they are stated doesn't seem clear to me.
tensor-products
asked Apr 9 at 3:08
HJoyHJoy
816
$endgroup$
I think I may just be confused by the question... would appreciate help!
I am given the function,
$f(ae_1 + be_2 , ce_1 + de_2) = ac + bd$
(it doesn't say, but I assume we have $e_1, e_2$ the standard basis for $R^2$)
First thing was to show that $f$ is bi-linear, which I have confirmed. Then I am to "expand the 2-tensor in the basis". Can someone explain what this means? Maybe I'm just not liking how it is stated. Meaning, write it as a sum $sum e_i otimes e_j$?
Secondly, show $e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = ac$.
This seems simple enough, as (my understanding is) $e_1$ "keeps" the 1st term.
$e_1 otimes e_1 (ae_1 + be_2 , ce_1 + de_2) = e_1(ae_1 + be_2)cdot e_2(ce_1 + de_2) = a cdot c$
Third, express $(ae_1 + be_2) wedge (ce_1 + de_2)$ in terms of $ac(e_1 times e_2) + bd(e_2 times e_2)$. This should be tensor product, right? Not cross product, as it is stated.
My professor gives us sets of notes and problems to work on before going to class. He writes the questions himself, so sometimes the way they are stated doesn't seem clear to me.
tensor-products
tensor-products
asked Apr 9 at 3:08
HJoyHJoy
816
asked Apr 9 at 3:08
HJoyHJoy
816
asked Apr 9 at 3:08
HJoyHJoy
816
asked Apr 9 at 3:08
HJoyHJoy
816
asked Apr 9 at 3:08
HJoyHJoy
816
816
add a comment |
add a comment |
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