Kronecker delta question

  • Thread starter quietrain
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  • #1
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does anyone know why

djmdkl = dkm

when j = l

thanks!

can i interchange it to djmdkl = dkldjm ?

since l = j
do the "j" in dkjdjm signify summation?

so i get dk1d1m + dk2d2m + dk3d3m ?
 

Answers and Replies

  • #2
Pengwuino
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Well, follow the logic the deltas imply.

If j=l, then the first delta function can be written [itex]\delta_{lm}[/itex] and you have [itex]\delta_{lm} \delta_{kl}[/itex]. This combination is only non-zero when l = m and l = k which implies k=m, so you can rewrite the whole thing as [itex]\delta_{km}[/itex]
 
  • #3
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oh i see thank you , so its by logic and inference?
 
  • #4
Pengwuino
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Yes, although formally a summation (not in the way you put it, I don't know how you claim there's a summation in what you wrote) does occur but it's fairly simple logic as to why it's true.
 
  • #5
HallsofIvy
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I woud say that "[itex]\delta_{jm}\delta_{kl}[/itex] with j= l" does not imply as sum but that [itex]\delta_{jm}\delta_{kj}[/itex] does, by the Einstein summation convention.
 
  • #6
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I woud say that "[itex]\delta_{jm}\delta_{kl}[/itex] with j= l" does not imply as sum but that [itex]\delta_{jm}\delta_{kj}[/itex] does, by the Einstein summation convention.
More to this is to think of the kronecker delta as a symmetric unit matrix and that allows you to write;
[itex]\delta_{jm}=\delta_{mj}=\delta_{jl}\delta_{lm}=\delta_{lm}\delta_{jl}[/itex]
 
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  • #7
Fredrik
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does anyone know why

djmdkl = dkm

when j = l
It's not. For example, [itex]\delta_{3m}\delta_{k3}[/itex] is not equal to [itex]\delta_{km}[/itex] in general. It is when m=k=3, but not when m=k=2.

On the other hand [itex]\delta_{jm}\delta_{kj}=\delta_{km}[/itex], because this is a sum with three terms, and in every term, the first factor is 0 if j≠m and the second factor is 0 if j≠k. So if k≠m, all the terms are =0 because they all contain at least one factor that's zero, but if k=m, the term with j=m=k is =1 and the others =0.

can i interchange it to djmdkl = dkldjm ?
Yes, because for each i and each j, [itex]\delta_{ij}[/itex] is a real number, and real numbers commute.

do the "j" in dkjdjm signify summation?
Yes.

so i get dk1d1m + dk2d2m + dk3d3m ?
Yes.
 
  • #8
Fredrik
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More to this is to think of the kronecker delta as a symmetric unit matrix and that allows you to write;
[itex]\delta_{jm}=\delta_{mj}=\delta_{jl}\delta_{lm} =\delta_{lm}\delta_{jl}[/itex]
You need to make sure that your latex code contains a space at least once every 50 characters. If if it doesn't, vBulletin will insert a space and break the code.
 
  • #9
Fredrik
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Quietrain, do you understand the answers you've been given? If not, then please say something. (Would you rather start over from square one seven months from now?)

since l = j
do the "j" in dkjdjm signify summation?

so i get dk1d1m + dk2d2m + dk3d3m ?
I didn't even notice what you said here, but you're making a serious error. It doesn't make sense to start this sentence with "since l=j". There's no l in the expression that follows, and the j that is a part of that expression is summed over, so it doesn't have a specific value. This means that no matter what l is, it wouldn't make any sense to say that l=j in that expression.

You are always extremely careless with your statements. I think you would find mathematics a lot easier if you made a habit of trying to express yourself as clearly as possible.

can i interchange it to djmdkl = dkldjm ?
This question proves that you either think that one of the equalities \begin{align}
0\cdot 0=0\cdot 0\\
0\cdot 1=1\cdot 0\\
1\cdot 0=0\cdot 1\\
1\cdot 1=1\cdot 1
\end{align} is false, or that you don't understand the notation that puts one or more indices on a symbol. I don't believe that you would doubt those equalities, so I have to conclude that you don't understand what any symbol with an index on it represents. If you want to understand any of these things, you have to start by making sure that you understand the basics.
 
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  • #10
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i think hallsofivy knows what i am trying to say,

maybe i should rephrase it


i have this

djmdkl

when j = l

does it give me

OPTION 1)

dkm since Pengwuino said that its done by logic deduction. but fredrik says that its not? he says that djmdkj = dkm ,
but if i let j=l , issn't djmdkl = djmdkj ??

OR

OPTION 2)

djmdkl

since l = j

djmdkj where J is now a dummy variable? with einstein summation of repeated indices,

so i get dk1d1m + dk2d2m + dk3d3m ?

hallsofivy says its not einstein summation? and fredrik says this is wrong?

I didn't even notice what you said here, but you're making a serious error. It doesn't make sense to start this sentence with "since l=j". There's no l in the expression that follows, and the j that is a part of that expression is summed over, so it doesn't have a specific value. This means that no matter what l is, it wouldn't make any sense to say that l=j in that expression.
are you trying to say that since J is now a dummy variable, i cannot let l = j ? or else l is no longer a free index?



CONCLUSION

so is dkm = dk1d1m + dk2d2m + dk3d3m ?? debrook's post seems to suggest this is right?

so when k=m, i get 1+1+1 = 3?

or are they 2 different entities?

but if i let k=m, dkk = 3 right? assuming k goes from 1 to 3

so d11+d22+d33 = 1+1+1=3?

so is d11+d22+d33 the same as dk1d1m + dk2d2m + dk3d3m ??? (they give same answer but do they mean the same thing?)

oh well, this is mind boggling ,,,
 
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  • #11
Fredrik
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but if i let j=l , issn't djmdkl = djmdkj ??
No. There's only one term on the left and three terms on the right. (Before we have actually checked, it's conceivable that the value of the expression on the left is the same as the value of the expression on the right for all values of the indices. But if you understand the notation at all, it's immediately obvious that the two expressions represent different calculations).

hallsofivy says its not einstein summation? and fredrik says this is wrong?
Halls and I said the same thing. Einstein's summation convention is that repeated indices are summed over. There's no repeated index in [itex]\delta_{jm}\delta_{kl}[/itex], but there is a repeated index in [itex]\delta_{jm}\delta_{kj}[/itex].

are you trying to say that since J is now a dummy variable, i cannot let l = j ?
Maybe I should have just said that it doesn't make sense to say "since l=j" and then ask a question that doesn't involve l. But another problem with the question we're discussing (quoted again below) is that it's a yes/no question, and if the answer is yes, then the question fails to make sense in a second way (by setting a variable that's summed over to a fixed value).

Just a reminder of what we're talking about:
since l = j
do the "j" in dkjdjm signify summation?

so is dkm = dk1d1m + dk2d2m + dk3d3m ??
Why don't you check this yourself by plugging in all possible values of k and m? There are only 9 combos. (The answer is "yes", and if you read my first post again, you'll see that I said this there).

so when k=m, i get 1+1+1 = 3?
Do you think a variable can represent different numbers at different places in the same expression? This would mean that an equality like x+x=2x is usually false.


but if i let k=m, dkk = 3 right? assuming k goes from 1 to 3
You're not making sense here. However, both of the following statements are true:
(a) [itex]\delta_{kk}=3[/itex].
(b) If [itex]k=m[/itex], then [itex]\delta_{km}=1[/itex].


so is d11+d22+d33 the same as dk1d1m + dk2d2m + dk3d3m ???
Again, a variable never represents different numbers at different places in the same expression. That would make it impossible to do mathematics.
 
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  • #12
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QUESTION 1)
oh.. so ,d11+d22+d33 is not the same as dk1d1m + dk2d2m + dk3d3m

because k or m cannot be 1,2,3 in the same expression?



Q 2)
for dkm = dk1d1m + dk2d2m + dk3d3m

so i only get d11 = d11d11 +0 + 0 = 1
so its kind of redundant to write the other two terms? unless i specify that k and m goes from 1 to 3 in dkm?

or does writing dkm already implies that k and m must go from 1 to 3?




Q 3)
with regards to the l=j part,

i was actually looking at this,
kroneckerdelta.jpg

(weird the image code is not working?)

kronecker delta.jpg


so after j=l, how did it become 3 dkm - dkm ?

3 dkm came from this part : djldkm right? while

- dkm came from the contraction of djmdkl ? but you said this contraction is wrong in your first post?

also, how did the 3 dkm part have a 3 popping out? shouldn't djldkm = 1dkm where l=j ?

or is the crux of this problem lying with the levi -civita symbol on the left? where there is actually a summation symbol for repeated indices going from 1to 3?

so that i get { d11+ d22 +d33 }dkm which is 3dkm ?



Q4)
also, i really don't understand this part. dkk is not equal to dkm when k=m as per your last post. you mean dkm where k=m, is 1, but i cannot write it as dkk as the repeated k here would mean having repeated indices,which clashes with dkk=3 when k goes from 1 to 3, so, only if the expression straight off the bat is dkk, then its summation?




Q5)
so, what does say, d12 tell me? i know if index are equal, it is =1. the index are not really rows and columns indicators right? then what are they? is there something i can visualise or a simple example? since they are related to the levi-civita symbol, then it means that they have to have a meaning right? like for example, is my perception of levi-civita symbol correct as below?

since (A X B)= EijkAjBk=Ci means I must sum j and k(repeated index) for the ith component of C, thus if i goes from 1 to 3, then it has 3 components and thus is not a scalar but a vector like what the cross product gives? so i can imagine i to be say x-component, y, and z for 1,2,3

but for the kronecker delta, what do the index tell me?

thanks!
 
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  • #13
Fredrik
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QUESTION 1)
oh.. so ,d11+d22+d33 is not the same as dk1d1m + dk2d2m + dk3d3m

because k or m cannot be 1,2,3 in the same expression?
Right. If k=m, the second expression can still represent three different calculations, but none of them is [itex]\delta_{11}\delta_{11} +\delta_{22}\delta_{22}+\delta_{33}\delta_{33}[/itex].

Q 2)
for dkm = dk1d1m + dk2d2m + dk3d3m

so i only get d11 = d11d11 +0 + 0 = 1
so its kind of redundant to write the other two terms? unless i specify that k and m goes from 1 to 3 in dkm?

or does writing dkm already implies that k and m must go from 1 to 3?
Your calculation is correct. I don't know what it would mean to say that "k and m goes from 1 to 3 in dkm".


Q 3)
with regards to the l=j part,

i was actually looking at this,
kroneckerdelta.jpg

(weird the image code is not working?)

View attachment 38623

so after j=l, how did it become 3 dkm - dkm ?

3 dkm came from this part : djldkm right? while

- dkm came from the contraction of djmdkl ? but you said this contraction is wrong in your first post?

also, how did the 3 dkm part have a 3 popping out? shouldn't djldkm = 1dkm where l=j ?

or is the crux of this problem lying with the levi -civita symbol on the left? where there is actually a summation symbol for repeated indices going from 1to 3?

so that i get { d11+ d22 +d33 }dkm which is 3dkm ?
I don't like how the author uses the phrase "setting [itex]j=l[/itex]". I would interpret that as "when j and l represent the same number, we get..." That's not at all what he's trying to say. What he meant to say is that the statement

[tex]\text{For all $j,k,l,m\in\{1,2,3\}$, we have } \epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}.[/tex] implies the statement [tex]\text{For all $k,m\in\{1,2,3\}$, we have }\epsilon_{ijk}\epsilon_{ijm}=\delta_{jj}\delta_{km}-\delta_{jm}\delta_{kj}=3\delta_{km}-\delta_{km}=2\delta_{km}.[/tex] You might want to read my first post again to see what I'm actually saying.

Q4)
also, i really don't understand this part. dkk is not equal to dkm when k=m as per your last post. you mean dkm where k=m, is 1, but i cannot write it as dkk as the repeated k here would mean having repeated indices,which clashes with dkk=3 when k goes from 1 to 3, so, only if the expression straight off the bat is dkk, then its summation?
Yes, that's exactly right.

Q5)
so, what does say, d12 tell me? i know if index are equal, it is =1. the index are not really rows and columns indicators right? then what are they? is there something i can visualise or a simple example? since they are related to the levi-civita symbol, then it means that they have to have a meaning right? like for example, is my perception of levi-civita symbol correct as below?

since (A X B)= EijkAjBk=Ci means I must sum j and k(repeated index) for the ith component of C, thus if i goes from 1 to 3, then it has 3 components and thus is not a scalar but a vector like what the cross product gives? so i can imagine i to be say x-component, y, and z for 1,2,3

but for the kronecker delta, what do the index tell me?

thanks!
[itex]\delta_{12}[/itex] is a symbol that that represents the number 0, nothing more, nothing less. There's nothing in the definition of [itex]\delta_{ij}[/itex] that says that i and j represents rows and columns, but for each value of i and j, [itex]\delta_{ij}[/itex] happens to be equal to the component on row i, column j, of the 3×3 identity matrix. So you can think of the indices as representing rows and columns if you want to. The indices are just variables that represent numbers in the set {1,2,3}. [itex]\delta[/itex] is a function from {1,2,3}×{1,2,3} into {0,1}. The fact that we write [itex]\delta_{ij}[/itex] instead of [itex]\delta(i,j)[/itex] is just a notational convention.
 
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  • #14
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alright i think i understand a lot now :O

its all about the basics.. like what my prof n you keep saying hahah

thanks!
 
  • #15
Fredrik
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I'm pleased that we've made progress. I think that now that we have found that one really important thing that you've been doing consistently wrong, you will find everything else much easier.
 
  • #16
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easier? perhaps

i found more questions though :(
 
  • #17
Fredrik
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Go ahead, ask them. Better to do it now than to wait until you've forgotten what you've learned in these threads.

I think it's best if you just ask one question at a time. I will be giving you hints rather than complete solutions from now on. When you ask a question, you should include your own attempt to answer it, up to the point where you get stuck.

I also have to request that when you write down a question, you think it through before you post it. You need to work harder to make sure that your questions make sense. It's too hard to try to figure out what you mean sometimes.
 
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  • #18
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ok this question is more about tensors



for example, my prof says x12 + x22 is an invariant

invariant means the tensor is order 0 right? means it has no index on it?

just purely X? not Xi or Xij etc , that is to say it has no components right?

so since x12 + x22 has indexes, why is it invariant?

also, what is an invariant? is it a scalar?
 
  • #19
Fredrik
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for example, my prof says x12 + x22 is an invariant
The expression [itex]x_1^2+x_2^2[/itex] is invariant under rotations in the 12 plane (the xy plane) in the sense that if you rotate the vector [itex](x_1,x_2)[/itex] in the 12 plane by some angle [itex]\theta[/itex] and the components of the new vector are [itex](x'_1,x'_2)[/itex], we have [itex]x'_1^2+x'_2^2=x_1^2+x_2^2[/itex], no matter what [itex]\theta[/itex] is.

This is how I prefer to think about these things (when I'm forced to think about it in terms of coordinates): The expression [itex]x_1^2+x_2^2[/itex] can be thought of as a formula that associates a real number with each coordinate system. This association is said to be invariant if it associates the same real number with each coordinate system. More generally, an association of an indexed set of numbers with each coordinate system is said to be a tensor if the indexed set associated with one coordinate system is related to the indexed set associated with another as described by the tensor transformation law.

(In this context, the only coordinate systems we're considering are related to each other by rotations).

invariant means the tensor is order 0 right? means it has no index on it?

just purely X? not Xi or Xij etc , that is to say it has no components right?
Right. The indices indicate which component we're dealing with, and scalars (=invariants) don't have components. Edit: I mean that a scalar has exactly one component, so there's no need to use an index to distinguish between different components.

so since x12 + x22 has indexes, why is it invariant?
Because just like any other expression that's constructed from tensors but doesn't have any free indices, like e.g. [itex]A_{ij}B_{ij}[/itex], it associates the same real number with each coordinate system. By a "free" index, I mean one that isn't summed over. [itex]x_1^2+x_2^2=x_i x_i[/itex] clearly doesn't have any free indices.
 
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  • #20
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The expression [itex]x_1^2+x_2^2[/itex] is invariant under rotations in the 12 plane (the xy plane) in the sense that if you rotate the vector [itex](x_1,x_2)[/itex] in the 12 plane by some angle [itex]\theta[/itex] and the components of the new vector are [itex](x'_1,x'_2)[/itex], we have [itex]x'_1^2+x'_2^2=x_1^2+x_2^2[/itex], no matter what [itex]\theta[/itex] is.
oh.. is there a way to "see" whether something is invariant or not? or do we have to manually work it out using the transformation law? i.e doing the matrix multiplication etc...

because my prof said [itex]x_1^2+x_2^2[/itex] is invariant and (i think he said this)[itex]x_1^3+x_2^3[/itex] is not invariant straight away.


also with regards to the quotient law of tensors that says

(A)(B) = C , if B,C are tensors, A is also a tensor ,if order of C = order of A + order of B - 2

so if i want to show that A is a 2nd order tensor, the idea to solving is just to mulitply it by a order 0 (scalar) so that i get order 0 (scalar)

so its something like this right

(A) (scalar) = (scalar , e.g 0 to make life easier)

so what is the idea to figuring out what (scalar) to mulitply A by?

because my prof gave me a 2nd order tensor (3x3 matrix) with crazy sin and cos terms everywhere
 
  • #21
Fredrik
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oh.. is there a way to "see" whether something is invariant or not? or do we have to manually work it out using the transformation law? i.e doing the matrix multiplication etc...

because my prof said [itex]x_1^2+x_2^2[/itex] is invariant and (i think he said this)[itex]x_1^3+x_2^3[/itex] is not invariant straight away.
If it's a combination of tensors, with no free indices, then it's always a scalar. If it's a combination of tensors, with at least one free index, then it's not a scalar. If it's not a combination of tensors at all, you will have to figure out how it associates numbers with coordinate systems.

Have you done the calculation that verifies that [itex]x_1^2+x_2^2[/itex] is invariant? If not, you should do it right away.

Matrix version: Since [itex]x'=Rx[/itex], we have [itex]x'^Tx'=\dots[/itex]

Component version: Since [itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i=\dots[/itex]

Can you fill in the rest?

also with regards to the quotient law of tensors that says

(A)(B) = C , if B,C are tensors, A is also a tensor ,if order of C = order of A + order of B - 2

so if i want to show that A is a 2nd order tensor, the idea to solving is just to mulitply it by a order 0 (scalar) so that i get order 0 (scalar)

so its something like this right

(A) (scalar) = (scalar , e.g 0 to make life easier)

so what is the idea to figuring out what (scalar) to mulitply A by?

because my prof gave me a 2nd order tensor (3x3 matrix) with crazy sin and cos terms everywhere
I'm actually not familiar with this rule. Can you state it exactly as it appears in your book or your lecture notes? (I assume that there are indices that you haven't included here).
 
  • #22
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it was kind of hard to find anything on quotient law on the internet... weird.

middle of page 41
http://www.hereticalcosmology.com/articles/Tensor_Analysis-Chapter_1.pdf [Broken]

this version is a lot more complicated though


below is the version that my lect gave me

if B and C are tensors and also that
Apq...k...mBij...k...n = Cpq...mij ...n
holds in all rotated coordinate frames where A, B, and C are of
Mth, Nth and (M+N-2)th order, then Apq...k...m is also a
tensor.




with regards to the invariant question

i have only learnt how to transform vectors like (x2 , -x1)

so that v1' = L11v1 + L12v2
so that v2' = L21v1 + L22v2

where my v1 takes the first vector x2
and my v2 takes the 2nd vector -x1



so with regards to the x12 + x22

what is my v1 and v2 ??

am i suppose to take v1 = x12 + x22 ?

then what will my v2 be?

or am i suppose to take v1 = x1
v2 = x2

so that by the transformation R,

x12' + x22' = { L11x1+ L12x2 } 2 +
L21x1+ L22x2 } 2

where c = cos, s = sin

= { cx1 + sx2 }2 + { -sx1 + cx2 }2 }

= x12 + x22


so in the component form

is it suppose to read like this

[itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i= R_{ij}x_j R_{ik}x_k[/itex]

[itex]x'_1 x'_1 + x'_2 x'_2 =[ R_{11}x_1 + R_{12} x_2 ][ R_{11}x_1 + R_{12}x_2] + [ R_{21}x_1 + R_{22} x_2 ][ R_{21}x_1 + R_{22}x_2][/itex]
 
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  • #23
Fredrik
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it was kind of hard to find anything on quotient law on the internet... weird.

middle of page 41
http://www.hereticalcosmology.com/articles/Tensor_Analysis-Chapter_1.pdf [Broken]
The proof on pages 41-42 is not too hard to understand, so you should work it through on your own. Ask if you get stuck on a detail. Note that it's essential that (13.5) holds for all tensors B, and in all the coordinate systems under consideration. If it just holds for one specific B, or in some coordinate systems, (13.5) doesn't really tell you anything.

I'm not sure what you need help with regarding the quotient law. You may have to post the specific problem along with your attempt to solve it.

with regards to the invariant question

i have only learnt how to transform vectors like (x2 , -x1)

so that v1' = L11v1 + L12v2
so that v2' = L21v1 + L22v2

where my v1 takes the first vector x2
and my v2 takes the 2nd vector -x1
I don't understand what you're saying. I understand that L is a linear transformation that takes [itex](v_1,v_2)[/itex] to [itex](v'_1,v'_2)[/itex], but then you lost me. I can't make sense of the phrase "the first vector x2".


so with regards to the x12 + x22

what is my v1 and v2 ??

am i suppose to take v1 = x12 + x22 ?
No, you're supposed to "transform" the vector [itex](x_1,x_2)[/itex] the same way you transformed [itex](v_1,v_2)[/itex] above.

or am i suppose to take v1 = x1
v2 = x2
Yes, that's one way of putting it.

so that by the transformation R,

x12' + x22' = { L11x1+ L12x2 } 2 +
L21x1+ L22x2 } 2

where c = cos, s = sin

= { cx1 + sx2 }2 + { -sx1 + cx2 }2 }

= x12 + x22
Correct.

so in the component form

is it suppose to read like this

[itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i= R_{ij}x_j R_{ik}x_k[/itex]

[itex]x'_1 x'_1 + x'_2 x'_2 =[ R_{11}x_1 + R_{12} x_2 ][ R_{11}x_1 + R_{12}x_2] + [ R_{21}x_1 + R_{22} x_2 ][ R_{21}x_1 + R_{22}x_2][/itex]
Yes, but you can simplify it without explicitly performing the summation.

[tex]x'_i x'_i=R_{ij}x_j R_{ik}x_k=x_j \underbrace{R_{ij}R_{ik}}_{=\delta_{jk}} x_k = x_j x_j[/tex]
It's actually much easier to do the whole thing with matrices instead of their components:
[tex]x'^Tx'=(Rx)^T(Rx)=x^TR^TRx=x^Tx[/tex]
 
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  • #24
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wow. so i did it through the brute force method :X , when i could have simplified it lol

with regards to the quotient law

if i have matrix A (3by3)

so if i mulitply it with a scalar B ( say 0) , then i get C = 0

then wouldn't it fulfill the quotient law?

since the orders of tensors are for A = 2 (M), B = 0 (N) , C = 0 (M+N-2)
 
  • #25
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What you're describing now has nothing to do with the quotient law, and it's not because of the orders of the tensors. Look at page 41 again. What the quotient law says about your components [itex]A_{ij}[/itex] and their relationship with a scalar B is this:
Suppose that for all real numbers B, [itex]A_{ij}B[/itex] are the components of a tensor. Then [itex]A_{ij}[/itex] are the components of a tensor.​
This is of course an entirely trivial statement.

What are you trying to do exactly? Are you trying to figure out if the numbers [itex]A_{ij}[/itex] are the components of a tensor? That wouldn't really make sense. You can always pick a coordinate system, and define A to be the tensor that has components [itex]A_{ij}[/itex] in that coordinate system. So no matter what numbers the [itex]A_{ij}[/itex] are, they are the components of infinitely many tensors. The question "Are these the components of a tensor?" only makes sense when what you've been given associates an indexed set of numbers with each coordinate system. Is there something about your matrix that gives you a reason to interpret it as an association of nine numbers with each coordinate system?

By the way, do you understand why [itex]R^TR=I[/itex], or equivalently, why [itex]R_{ij}R_{ik}=\delta_{jk}[/itex]? Do you understand that those two statements are actually the same statement written in different notations?
 
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