Changing the subject of an equation involving summation

[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

 

[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.

 

[resurgance2001]

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

 

[HallsofIvy]

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.

 

[Fredrik]

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 $(A^{-1})_{\alpha\beta}$.

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

 

[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^lSumming 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

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 $AB^T=C$, 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 $(A^{-1})_{\alpha\beta}$.

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

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 $AB^T=C$, 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 $(A^{-1})_{\alpha\beta}$.

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

Thanks - that is very clear

 

[Fredrik]

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 $(AB^T)^\beta{}_\gamma$.

 

[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

 

[resurgance2001]

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

 

[Fredrik]

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

Yes. Since the definition of matrix multiplication is $(AB)_{ij}=\sum_k A_{ik}B_{kj}$, we have $\sum_k A_{ki}B_{kj}=\sum_k(A^T)_{ik}B_{kj}=(A^TB)_{ij}$.