# Changing the subject of an equation involving summation

1. Oct 2, 2014

### resurgance2001

Hi

Can I ask a question please. I have an equation that involves the summation over some indices, for example.

A^αβ B_αγ = C^β_γ

Say that I don't know Β_αγ , and want to make this the subject of the equation, how is this done?

Thanks

Peter

2. Oct 2, 2014

### Ben Niehoff

On the left you have matrix multiplication written using index notation. So to isolate B, you need the matrix inverse to A, if it exists.

3. Oct 3, 2014

### resurgance2001

Thanks - that's what I thought it was but I just wanted to check.

4. Oct 3, 2014

### HallsofIvy

Staff Emeritus
Just to point out- the quantities in the original equation are NOT matrices, they are tensors. Of course, in a given coordinate system, they can be represented as matrices and then you can multiply both sides by the inverse matrix (assuming the matrix is invertible). By the definition of "tensor", while the individual numbers will vary from one coordinate system to another, the equation will still be valid and the solution matrix you get in one coordinate system will give the same tensor as will the solution matrix in any coordinate system.

5. Oct 4, 2014

### Fredrik

Staff Emeritus
The definition of matrix multiplication is $(AB)_{ij}=\sum_{k} A_{ik}B_{kj}$. If you compare this to the left-hand side of your $A^{\alpha\beta}B_{\alpha\gamma}=C^\beta{}_\gamma$, you see that your equation can be interpreted as row $\beta$, column $\gamma$ of a matrix equation $AB^T=C$ (Edit: See the correction in my next post), where A is the matrix with $A^{\alpha\beta}$ on row $\alpha$, column $\beta$, B is the matrix with $B_{\alpha\beta}$ on row $\alpha$, column $\beta$, and C is the matrix with $C^\alpha{}_\beta$ on row $\alpha$, column $\beta$. So if this A is an invertible matrix, then you can solve for $B^T$ by multiplying both sides of the matrix equation from the left by the inverse of A. If you want this to look good when you write out all the indices, you may want to use the convention that row $\alpha$, column $\beta$ of $A^{-1}$ is written as $(A^{-1})_{\alpha\beta}$.

$(A^{-1})_{\beta\delta}C^\beta{}_\gamma =(A^{-1})_{\beta\delta}A^{\alpha\beta}B_{\alpha\gamma} =(AA^{-1})_\delta^\alpha B_{\alpha\gamma} =\delta^\alpha_\delta B_{\alpha\gamma} =B_{\delta\gamma}$

Last edited: Oct 4, 2014
6. Oct 4, 2014

### resurgance2001

The actual problem I am working on a bit more complicated. I don't know if I can type it here clearly:

We are given:

Σ Γ^l_ik e_l . e_j + e_i . ΣΓ^l_jk e_l = δg_il / δx^l

Summing over l

I want to find the steps to rearrange this and make:

Γ^i_jk = .... ?

The answer is the standard equation giving the connection coefficients.

What I am trying to find is the steps that lead from ten
Thanks - that is very clear

7. Oct 4, 2014

### Fredrik

Staff Emeritus
I see now that I made a mistake. $A^{\alpha\beta}B_{\alpha\gamma}$ should be interpreted as $(A^TB)^\beta{}_\gamma$ or $(B^TA)_\gamma{}^\beta$, not as $(AB^T)^\beta{}_\gamma$.

8. Oct 4, 2014

### resurgance2001

Ok - what I would like to find is an interactive website which gives you lots of problems to work on for practice and then gives the correct answer and explanation. Does anyone know of a website like that?

Cheers

9. Oct 4, 2014

### resurgance2001

When you use say B^T, do you mean the transpose of B?

10. Oct 4, 2014

### Fredrik

Staff Emeritus
Yes. Since the definition of matrix multiplication is $(AB)_{ij}=\sum_k A_{ik}B_{kj}$, we have $\sum_k A_{ki}B_{kj}=\sum_k(A^T)_{ik}B_{kj}=(A^TB)_{ij}$.
The content structure of the passage is a series of posts from a forum where a user asks a question about manipulating tensor equations, and other users provide answers and further explanations.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Changing the subject of an equation involving summation

1. Oct 2, 2014

### resurgance2001

Hi

Can I ask a question please. I have an equation that involves the summation over some indices, for example.

A^αβ B_αγ = C^β_γ

Say that I don't know Β_αγ , and want to make this the subject of the equation, how is this done?

Thanks

Peter

2. Oct 2, 2014

### Ben Niehoff

On the left you have matrix multiplication written using index notation. So to isolate B, you need the matrix inverse to A, if it exists.

3. Oct 3, 2014

### resurgance2001

Thanks - that's what I thought it was but I just wanted to check.

4. Oct 3, 2014

### HallsofIvy

Staff Emeritus
Just to point out- the quantities in the original equation are NOT matrices, they are tensors. Of course, in a given coordinate system, they can be represented as matrices and then you can multiply both sides by the inverse matrix (assuming the matrix is invertible). By the definition of "tensor", while the individual numbers will vary from one coordinate system to another, the equation will still be valid and the solution matrix you get in one coordinate system will give the same tensor as will the solution matrix in any coordinate system.

5. Oct 4, 2014

### Fredrik

Staff Emeritus
The definition of matrix multiplication is $(AB)_{ij}=\sum_{k} A_{ik}B_{kj}$. If you compare this to the left-hand side of your $A^{\alpha\beta}B_{\alpha\gamma}=C^\beta{}_\gamma$, you see that your equation can be interpreted as row $\beta$, column $\gamma$ of a matrix equation $AB^T=C$ (Edit: See the correction in my next post), where A is the matrix with $A^{\alpha\beta}$ on row $\alpha$, column $\beta$, B is the matrix with $B_{\alpha\beta}$ on row $\alpha$, column $\beta$, and C is the matrix with $C^\alpha{}_\beta$ on row $\alpha$, column $\beta$. So if this A is an invertible matrix, then you can solve for $B^T$ by multiplying both sides of the matrix equation from the left by the inverse of A. If you want this to look good when you write out all the indices, you may want to use the convention that row $\alpha$, column $\beta$ of $A^{-1}$ is written as $(A^{-1})_{\alpha\beta}$.

$(A^{-1})_{\beta\delta}C^\beta{}_\gamma =(A^{-1})_{\beta\delta}A^{\alpha\beta}B_{\alpha\gamma} =(AA^{-1})_\delta^\alpha B_{\alpha\gamma} =\delta^\alpha_\delta B_{\alpha\gamma} =B_{\delta\gamma}$

Last edited: Oct 4, 2014
6. Oct 4, 2014

### resurgance2001

The actual problem I am working on a bit more complicated. I don't know if I can type it here clearly:

We are given:

Σ Γ^l_ik e_l . e_j + e_i . ΣΓ^l_jk e_l = δg_il / δx^l

Summing over l

I want to find the steps to rearrange this and make:

Γ^i_jk = .... ?

The answer is the standard equation giving the connection coefficients.

What I am trying to find is the steps that lead from ten
Thanks - that is very clear

7. Oct 4, 2014

### Fredrik

Staff Emeritus
I see now that I made a mistake. $A^{\alpha\beta}B_{\alpha\gamma}$ should be interpreted as $(A^TB)^\beta{}_\gamma$ or $(B^TA)_\gamma{}^\beta$, not as $(AB^T)^\beta{}_\gamma$.

8. Oct 4, 2014

### resurgance2001

Ok - what I would like to find is an interactive website which gives you lots of problems to work on for practice and then gives the correct answer and explanation. Does anyone know of a website like that?

Cheers

9. Oct 4, 2014

### resurgance2001

When you use say B^T, do you mean the transpose of B?

10. Oct 4, 2014

### Fredrik

Staff Emeritus
Yes. Since the definition of matrix multiplication is $(AB)_{ij}=\sum_k A_{ik}B_{kj}$, we have $\sum_k A_{ki}B_{kj}=\sum_k(A^T)_{ik}B_{kj}=(A^TB)_{ij}$.
user
The content structure of the passage is a series of posts from a forum where a user asks a question about manipulating tensor equations, and other users provide answers and further explanations.

TYPE: SINGLE_QUESTION_WITH_ANSWER