text: # 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 $\sum_k A_{ki}B_{kj}=\sum_k(A^T)_{ik}B_{kj}=(A^TB)_{ij}$.
explanation: The document is a series of posts from a forum or discussion board where a user asks a question about manipulating an equation involving tensors and summation. The user seeks to isolate a tensor in the equation. Several responses are provided, offering explanations and guidance on how to solve the problem. The discussion also touches on the distinction between tensors and matrices and the use of matrix inverses and transposes.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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 $\sum_k A_{ki}B_{kj}=\sum_k(A^T)_{ik}B_{kj}=(A^TB)_{ij}$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a series of posts from a forum or discussion board where a user asks a question about manipulating an equation involving tensors and summation. The user seeks to isolate a tensor in the equation. Several responses are provided, offering explanations and guidance on how to solve the problem. The discussion also touches on the distinction between tensors and matrices and the use of matrix inverses and transposes.
  
  Label: question-and-answer