Matrix Component Equation: Solving for B with Known Scalar and Indices

[member 428835]

Solve $$A_{ij} = c B_{ij}+B_{kk} \delta_{ij}$$ for $B$ where $c$ is a known scalar and $i,j,k$ are indices and range either $1,2,3$ and $\delta_{ij}$ which is the Kronecker Delta..

I've thought to write this into a matrix but I'm unsure what to do with the $B_{kk}$. Any help or guidance is greatly appreciated.

(This is from a book, so not exactly homework Thanks!

Josh

 

[Lucas SV]

One of the very useful techniques to use when solving such matrix equations is to take the trace of the equation and then substitute the result back in. I am assuming the Einstein summation convention in being applied so that $B_{kk}$ actually means $B_{11}+B_{22}+B_{33}$. Taking the trace yields $A_{ii}=(c+3)B_{kk}$, so you just substitute $B_{kk}$ back in and you are done.

This technique is very useful, so hopefully you will remember it to apply to other problems.

 

[member 428835]

Thanks for replying Lucas! So are you saying to rewrite the equation as $$A_{ij} = c B_{ij} + \frac{c}{c+3} A_{kk} \delta_{ij} \implies \\ B_{ij} = \frac{1}{c}A_{ij} - \frac{1}{c+3} A_{kk}\delta_{ij}$$ This totally makes sense! So that I understand, you use this trick whenever you have $X_{ii}$, right? Also, do you have any other fancy tricks? :)

 

[Lucas SV]

In the first line it is $\frac{1}{c+3}$ instead of what you wrote. Apart form that it is correct. Also I just realized that $c+3$ must be different than $0$ for this to work, otherwise you find $A$ to be traceless.

This totally makes sense! So that I understand, you use this trick whenever you have XiiXiiX_{ii}, right?

Not necessarely, although that may be a strong indicator to use it. Use it whenever you want a way to reduce the degrees of freedom of the equation. When taking the trace you reduce the freedom from $n^2$ to $1$ for $n$ by $n$ matrices.

In general a matrix will have $n^2$ components, and it can always be separated into a symmetric and antisymmetric part. The symmetric has $n(n+1)/2$ independent components while an antisymmetric part has $n(n-1)/2$ independent components. The symmetric part can be further separated into traceless part and trace. So taking the trace is analogous to computing a component of a matrix equation. You can also symmetrize or antisymmetrize to reduce the number of equations.

Another very useful technique is diagonlising one of the matrices. This can help simplify a particular problem.

 

[member 428835]

Thanks for catching the algebra mistake, and thanks for all your help!