# Transforming a full $4\times4$ symbolic matrix into a full $3\times3$ matrix and an eigenvalue

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

$$$$A'=U^{-1}A U=\begin{pmatrix} F_{3 \times 3} & 0_{3\times 1} \\ 0_{1 \times 3} & C \end{pmatrix}.$$$$ where $$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?

• What do you mean by "rewrite" ?
– A.G.
Commented Oct 29, 2020 at 1:48
• I meant transforming matrix A into A'. I have edited my title to clarify this point. Commented Oct 29, 2020 at 6:19
• ArrayPad[Array[a,{3,3}],{0,1}]+DiagonalMatrix[{0,0,0,b}] // TableForm
– I.M.
Commented Oct 29, 2020 at 7:56
• This is not a transformation. Commented Oct 29, 2020 at 8:01
• 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. Commented Oct 29, 2020 at 10:49

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]


• 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. Commented Oct 29, 2020 at 7:42
• A diagonal Matrix is a special case of a full matrix Commented Oct 29, 2020 at 9:17
• Indeed! But that is not what I am looking for. Commented Oct 29, 2020 at 9:37