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Give this matrix A: \begin{pmatrix}-25&2&3&-29\\2&7&7&11\\3&7&7&2\\-29&11&2&11\end{pmatrix}

How can we calculate C matrix when A = AC - CA without extensive computations? Thought of doing these steps:

  1. Let C be {{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}}
  2. Calculate (via wolfram) AC, CA
  3. Subtract AC, CA (cannot do it with wolfram, there's a limit on the characters of input
  4. Then assign the 1st row of the result with the respective row of matrix A
  5. Solve the system of 4 linear equations with 4 variables

Is that correct? How can I find a workaround for step 3?

Thank you for your time!

  • 0
    Why only the look at the first row of the equality A=AC-CA? Wouldn't you need equality in all four rows? Also: wouldn't C be unique only up to addition by matrices B such that AB-BA=0 (ie, the kernel of the adjoint, $\operatorname{ker}\,\operatorname{ad}_A$)?2012-05-17
  • 0
    @anon: "Solve the system of 4 linear equations with 4 variables", that is what I mean, all the 4 rows (I only mentioned the 1st row).2012-05-17

2 Answers 2