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I have a $4\times4$ matrix

A=\begin{pmatrix} 0.16 (\cos (\text{kx})+2) & 0.55 \cos \left(\frac{\text{kx}}{2}\right)+(0.\, +0.76 i) \sin \left(\frac{\text{kx}}{2}\right) & 0.55 \cos \left(\frac{\text{kx}}{2}\right) & 0.55\\ 0.55 \cos \left(\frac{\text{kx}}{2}\right)-(0.\, +0.76 i) \sin \left(\frac{\text{kx}}{2}\right) & 0.16 (\cos (\text{kx})+2) & 0.55 & 0.55 \cos \left(\frac{\text{kx}}{2}\right) \\ 0.55 \cos \left(\frac{\text{kx}}{2}\right) & 0.55& 0.16 (\cos (\text{kx})+2) & 0.55 \cos \left(\frac{\text{kx}}{2}\right)+(0.\, +0.76 i) \sin \left(\frac{\text{kx}}{2}\right) \\ 0.55 & 0.55 \cos \left(\frac{\text{kx}}{2}\right) & 0.55 \cos \left(\frac{\text{kx}}{2}\right)-(0.\, +0.76 i) \sin \left(\frac{\text{kx}}{2}\right) & 0.16 (\cos (\text{kx})+2) \\ \end{pmatrix}

A={{0.16*(2 + Cos[kx]), 0.55*Cos[kx/2] + (a + 0.76*I)*Sin[kx/2], 
      0.55*Cos[kx/2], 
  0.55 }, {0.55*Cos[kx/2] - (a + 0.76*I)*Sin[kx/2], 
      0.16*(2 + Cos[kx]), 0.55 , 0.55*Cos[kx/2]}, 
    {0.55*Cos[kx/2], 0.55, 0.16*(2 + Cos[kx]), 
      0.55*Cos[kx/2] + (a + 0.76*I)*Sin[kx/2]}, {0.55 , 
  0.55*Cos[kx/2], 
      0.55*Cos[kx/2] - (a + 0.76*I)*Sin[kx/2], 0.16*(2 + Cos[kx])}}

where $a$ and $k_x$ are real.

I would like to transform this matrix into a full $3 \times 3$ matrix $F_{3\times 3}(k_x,a)$ and a diagonal term $C(k_x,a)$, where $C$ is one of the eigenvalues of A, as

\begin{equation} A'=U^{-1}A U=\begin{pmatrix} F_{3 \times 3} & 0_{3\times 1} \\ 0_{1 \times 3} & C \end{pmatrix}. \end{equation} where $U$ is the transforming matrix. When $U$ is a matrix with eigenvectors on its coloumns, the above transformation simply diagonalizes the full matrix and $F_{3 \times 3}$ will be diagonal. This particular $U$ is not what I am seeking, as I would like $F_{3\times3}$ to be a full matrix with non-diagonal elements. In principle, I can get this form of $A'$ using linear combinations of rows and columns. How can I perform such a transformation on A using eigenvectors of A?

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    $\begingroup$ What do you mean by "rewrite" ? $\endgroup$
    – A.G.
    Oct 29, 2020 at 1:48
  • $\begingroup$ I meant transforming matrix A into A'. I have edited my title to clarify this point. $\endgroup$
    – Shasa
    Oct 29, 2020 at 6:19
  • $\begingroup$ ArrayPad[Array[a,{3,3}],{0,1}]+DiagonalMatrix[{0,0,0,b}] // TableForm $\endgroup$
    – I.M.
    Oct 29, 2020 at 7:56
  • $\begingroup$ This is not a transformation. $\endgroup$
    – Shasa
    Oct 29, 2020 at 8:01
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    $\begingroup$ First you diagonalize A to A'. Then you leave the forth eigenvector untouched and make a base change among eigenvectors 1..3. This will give you a full F matrix. But as see, this base change is arbitrary, there are infinite many solutions. $\endgroup$ Oct 29, 2020 at 10:49

1 Answer 1

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Try

  A = A /. kx -> k x

Then

      DiagonalMatrix[
 Eigenvalues[A //. Cos[k x] -> ck /. Cos[k x/2] -> ck2] //. 
    Cos[k x] -> ck /. Cos[k x/2] -> ck2 // FullSimplify]

enter image description here

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  • $\begingroup$ Well, I do not want to have a diagonal matrix. It is important to me that F_{3\times3} be a full matrix and not a diagonal one. $\endgroup$
    – Shasa
    Oct 29, 2020 at 7:42
  • $\begingroup$ A diagonal Matrix is a special case of a full matrix $\endgroup$
    – chris
    Oct 29, 2020 at 9:17
  • $\begingroup$ Indeed! But that is not what I am looking for. $\endgroup$
    – Shasa
    Oct 29, 2020 at 9:37

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