Prove that K is a tensor using quotient theorem

[MatinSAR]

Homework Statement

Prove that $K_{ij}$ is a tensor using quotient theorem.

Relevant Equations

Tensor analysis and quotient theorem.

$$K'_{ij}A'_{jk}=B'_{ik}=a_{ip}a_{kq}B_{pq}=a_{ip}a_{kq}K_{pr}A_{rq}=a_{ip}a_{kq}K_{pr}a_{jr}a_{kq}A'_{jk}$$$$K'_{ij}=a_{ip}a_{kq}a_{kq}a_{jr}K_{pr}$$

Can someone point out my mistake? What I've found shows that K is not a tensor.
It is different from my book and I cannot find my mistake. According to book K should be a 2nd-rank tensor.

 

[Hill]

You use k for a dummy/summation index on the right, but it is already used as a free index on the left.

 

[MatinSAR]

You use k for a dummy/summation index on the right, but it is already used as a free index on the left.

In last step I want to write $A_{rq}$ in prime corrdinates. So I had to use "k" again ... I wanted to prove using this method which is used by the book. Does this method work for my question in post #1?

 

[Hill]

So I had to use "k" again

But you can't. "k" is fixed by the left side of the equation, so you can't "sum for all k" on the right side of the equation.

Does this method work for my question in post #1?

I'd rather convert the $K_{ij}A_{jk}=B_{ik}$ into a $X_{ijk}C_{j}=B_{ik}$ and apply the quotient rule you already have.

 

[MatinSAR]

But you can't. "k" is fixed by the left side of the equation, so you can't "sum for all k" on the right side of the equation.

I'd rather convert the $K_{ij}A_{jk}=B_{ik}$ into a $X_{ijk}C_{j}=B_{ik}$ and apply the quotient rule you already have.

Then I will find out what is $X$. But questions asks for $K$.

 

[MatinSAR]

$$X'_{ijk}C'_j=B'_{ik}=a_{ip}a_{kq}B_{pq}=a_{ip}a_{kq}X_{prq}C_r$$ $$X'_{ijk}C'_j=a_{ip}a_{kq}X_{prq}a_{jr}C'_j$$$$X'_{ijk}=a_{ip}a_{jr}a_{kq}X_{prq}$$

I've proved it's a 3rd-rank tensor. But still I don't see how it helps me to prove that $K$ is a 2nd-rank tensor.

 

[Hill]

$$X'_{ijk}C'_j=B'_{ik}=a_{ip}a_{kq}B_{pq}=a_{ip}a_{kq}X_{prq}C_r$$ $$X'_{ijk}C'_j=a_{ip}a_{kq}X_{prq}a_{jr}C'_j$$$$X'_{ijk}=a_{ip}a_{jr}a_{kq}X_{prq}$$

I've proved it's a 3rd-rank tensor. But still I don't see how it helps me to prove that $K$ is a 2nd-rank tensor.

Very well. Now, back to $K_{ij}A_{jk}=B_{ik}$.
Since $A_{jk}$ is arbitrary, we can take it as $A_{jk}=C_j D_k$. Then we have $K_{ij}C_j D_k=B_{ik}$.
Let $X_{ijk}=K_{ij}D_k$. Then $X_{ijk}C_j=B_{ik}$. You've shown that it makes $X_{ijk}$ a tensor.
So, now you have $K_{ij}D_k=X_{ijk}$. Show that $K_{ij}$ is a tensor.

 

[MatinSAR]

Very well. Now, back to $K_{ij}A_{jk}=B_{ik}$.
Since $A_{jk}$ is arbitrary, we can take it as $A_{jk}=C_j D_k$. Then we have $K_{ij}C_j D_k=B_{ik}$.
Let $X_{ijk}=K_{ij}D_k$. Then $X_{ijk}C_j=B_{ik}$. You've shown that it makes $X_{ijk}$ a tensor.
So, now you have $K_{ij}D_k=X_{ijk}$. Show that $K_{ij}$ is a tensor.

Great! What a clever idea! Thanks a lot for your help @Hill ...

I'll try. I hope I will be able to prove.

 

[MatinSAR]

@Hill
Sadly I wasn't able to prove for the same reason which was mentioned is post #1.
$$K'_{ij}D'_k=X'_{ijk}=a_{ip}a_{jq}a_{kr}X_{pqr}=a_{ip}a_{jq}a_{kr}K_{pq}D_{r}=a_{ip}a_{jq}a_{kr}K_{pq}a_{kr}D'_k$$$$K'_{ij}=a_{ip}a_{jq}a_{kr}a_{kr}K_{pq}$$

I am trying to use my book's method. This method worked for equation in post #6 but It doesn't work for equation in post #1 and it doesn't work in this post as well ... I dont know why !

 

[Hill]

@Hill
Sadly I wasn't able to prove for the same reason which was mentioned is post #1.
$$K'_{ij}D'_k=X'_{ijk}=a_{ip}a_{jq}a_{kr}X_{pqr}=a_{ip}a_{jq}a_{kr}K_{pq}D_{r}=a_{ip}a_{jq}a_{kr}K_{pq}a_{kr}D'_k$$$$K'_{ij}=a_{ip}a_{jq}a_{kr}a_{kr}K_{pq}$$

I am trying to use my book's method. This method worked for equation in post #6 but It doesn't work for equation in post #1 and it doesn't work in this post as well ... I dont know why !

$K'_{ij}D'_k=X'_{ijk}=a_{ip}a_{jq}a_{kr}X_{pqr}=a_{ip}a_{jq}a_{kr}K_{pq}D_{r}=a_{ip}a_{jq}K_{pq}a_{kr}D_{r}=a_{ip}a_{jq}K_{pq}D'_k$

 

[MatinSAR]

$K'_{ij}D'_k=X'_{ijk}=a_{ip}a_{jq}a_{kr}X_{pqr}=a_{ip}a_{jq}a_{kr}K_{pq}D_{r}=a_{ip}a_{jq}K_{pq}a_{kr}D_{r}=a_{ip}a_{jq}K_{pq}D'_k$

I've found my mistake. Thanks for your help @Hill ;)