Calculating Eigenvalues using xAct

I'm trying to calculate the eigenvalues of the following system which originates from here (Appendix B): \begin{align} i{\xi}_{0}\hat{\tilde{\gamma}}_{ij} &= 2i\tilde{\gamma}_{k(i}\xi_{j)}\hat{\beta}^{k} - 2\alpha \hat{\tilde{A}}_{ij} - \frac{2i}{3}\tilde{\gamma}_{ij}\xi_{k}\hat{\beta}^{k}\,, \\ i{\xi}_{0}\hat{\chi}&=\frac{2}{3}\chi\left(\alpha\hat{K}-i\xi_{k}\hat{\beta}^{k}\right) \, , \\ i{\xi}_{0}\hat{\phi} &= -\alpha \hat{K}_{\phi} \,, \\ i{\xi}_{0}\hat{K} &= \hat{\alpha} + \frac{3b(x)\alpha}{4(1+b(x))}\left(\frac{\xi^{l}\xi^{k}\hat{\tilde{\gamma}}_{kl}}{\chi} - \frac{\tilde{\gamma}^{jk}\hat{\tilde{\gamma}}_{jk}}{2}\right) + i\alpha\chi\xi_{i}\hat{\hat{\Gamma}}^{i} - \left(\frac{2}{\chi}\hat{\chi}-\frac{\tilde{\gamma}^{ij}\hat{\tilde{\gamma}}_{ij}}{2}\right)\frac{\alpha(4+b(x))}{4(1+b(x))}\,, \\ i{\xi}_{0}\hat{K}_{\phi} &= - \alpha\hat{\phi}\,, \\ i{\xi}_{0}\hat{\Theta} &= - \frac{\alpha}{2(1+b(x))}\left(\frac{2}{\chi}\hat{\chi}-\frac{\tilde{\gamma}^{ij}\hat{\tilde{\gamma}}_{ij}}{2}\right) + \frac{i\alpha\chi\xi_{i}\hat{\hat{\Gamma}}^{i}}{2} + \frac{b(x)\alpha}{2(1+b(x))}\left(\frac{\xi^{l}\xi^{k}\hat{\tilde{\gamma}}_{kl}}{\chi} - \frac{\tilde{\gamma}^{jk}\hat{\tilde{\gamma}}_{jk}}{2}\right) \,,\\ i{\xi}_{0}\hat{\tilde{A}}_{ij} &= \left(\xi_{i}\xi_{j}-\frac{1}{3}\frac{\tilde{\gamma}_{ij}}{\chi}\right)\left(\chi\hat{\alpha}-\frac{\alpha}{2}\hat{\chi}\right) + i\alpha\chi\left(\tilde{\gamma}_{k(i}\xi_{j)}\hat{\hat{\Gamma}}^{k} - \frac{\tilde{\gamma}_{ij}\xi_{k}\hat{\hat{\Gamma}}^{k}}{3}\right) + \frac{1}{2}\alpha\left(\hat{\tilde{\gamma}}_{ij}-\frac{\tilde{\gamma}_{ij}\tilde{\gamma}^{kl}\hat{\tilde{\gamma}}_{kl}}{3}\right) \,,\\ i{\xi}_{0}\hat{\hat{\Gamma}}^{i} &= \frac{i\alpha}{\chi(1+b(x))}\left(-\frac{4}{3}\xi^{i}\hat{K}-2\chi b(x)\xi_{j}\hat{\tilde{A}}^{ij}+2\xi^{i}\hat{\Theta}\right) - \frac{1}{\chi}\left(\hat{\beta}^{i}+\frac{1}{3}\xi^{i}\xi_{l}\hat{\beta}^{l}\right) \,,\\ i{\xi}_{0}\hat{\alpha} &= -\frac{2\alpha}{1+a(x)}\left(\hat{K}-2\hat{\Theta}\right) \,,\\ i{\xi}_{0}\hat{\beta}^{i} &= \frac{3}{4(1+a(x))}\hat{\hat{\Gamma}}^{i} - \frac{ia(x)}{1+a(x)}\alpha\xi^{i}\hat{\alpha} \,. \end{align} The right hand side gives in matrix notation: \begin{equation} \begin{pmatrix} 0&\hat{\tilde{\gamma}}_{ef} & \hat{\chi} & \hat{\tilde{A}}_{cd} & \hat{K} & \hat{\Theta} & \hat{\hat{\Gamma}}^{g} & \hat{\alpha} & \hat{\beta}^{h} & \hat{\phi} & \hat{K}_{\phi} \\ \hat{\tilde{\gamma}}_{ij} & 0 & 0 & -2\alpha\delta_{i}^{c}\delta_{j}^{d} & 0 & 0 & 0 & 0 & 2i\tilde{\gamma}_{h(i}\xi_{j)}-\frac{2i}{3}\tilde{\gamma}_{ij}\xi_{h} & 0 & 0 \\ \hat{\chi} & 0 & 0 & 0 & \frac{2}{3}\chi\alpha & 0 & 0 & 0 & -\frac{2i}{3}\chi\xi_{h} & 0 & 0 \\ \hat{\tilde{A}}_{ab} & \frac{1}{2}\alpha\left(\delta_{a}^{e}\delta_{b}^{f}-\frac{\tilde{\gamma}_{ab}\tilde{\gamma}^{ef}}{3}\right) & -\frac{\alpha}{2}(\xi_{a}\xi_{b}-\frac{1}{3}\frac{\tilde{\gamma}_{ab}}{\chi}) & 0 & 0 & 0 & i\alpha\chi\left(\tilde{\gamma}_{g(a}\xi_{b)}-\frac{\tilde{\gamma}_{ab}\xi_{g}}{3}\right) & \chi(\xi_{a}\xi_{b}-\frac{1}{3}\frac{\tilde{\gamma}_{ab}}{\chi}) & 0 & 0 & 0 \\ \hat{K} & \frac{3b(x)\alpha}{4(1+b(x))}\left(\frac{\xi^{e}\xi^{f}}{\chi}-\frac{\tilde{\gamma}^{ef}}{2}\right) + \frac{\tilde{\gamma}^{ef}}{2}\frac{\alpha(4+b(x))}{4(1+b(x))} & -\frac{2}{\chi}\frac{\alpha(4+b(x))}{4(1+b(x))} & 0 & 0 & 0 & i\alpha\chi\xi_{g} & 1 & 0 & 0 & 0 \\ \hat{\Theta} & \frac{\alpha}{2(1+b(x))}\frac{\tilde{\gamma}^{ef}}{2} + \frac{b(x)\alpha}{2(1+b(x))}\left(\frac{\xi^{e}\xi^{f}}{\chi}-\frac{\tilde{\gamma}^{ef}}{2}\right) & -\frac{\alpha}{2(1+b(x))}\frac{2}{\chi} & 0 & 0 & 0 & \frac{i\alpha\chi\xi_{g}}{2} & 0 & 0 & 0 & 0 \\ \hat{\hat{\Gamma}}^{k} & 0 & 0 & \frac{i\alpha}{\chi(1+b(x))}(-2\chi b(x)\xi_{d}\delta_{c}^{k}) & \frac{i\alpha}{\chi(1+b(x))}\frac{-4\xi^{k}}{3} & \frac{i\alpha}{\chi(1+b(x))}2\xi^{k} & 0 & 0 & -\frac{1}{\chi}\left(\delta^{k}_{h} + \frac{\xi^{k}\xi_{h}}{3}\right) & 0 & 0 \\ \hat{\alpha} & 0 & 0 & 0 & -\frac{2\alpha}{1+a(x)} & \frac{4\alpha}{1+a(x)} & 0 & 0 & 0 & 0 & 0 \\ \hat{\beta}^{l} & 0 & 0 & 0 & 0 & 0 & \frac{3}{4(1+a(x))}\delta^{l}_{g} & -\frac{ia(x)}{1+a(x)}\alpha\xi^{l} & 0 & 0 & 0 \\ \hat{\phi} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\alpha \\ \hat{K}_{\phi} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\alpha & 0 \end{pmatrix} \end{equation} I typed the variables such that it is easier to read. $\tilde{\gamma}_{ij}$ is the conformal metric with $\chi$ the conformal factor, $\xi_{i}$ is a arbitrary vector with unit vector with respect to the induced three metric $\gamma_{ij}$.

However, if I feed this matrix to Mathematica using the xAct package, I get an error saying that it found inhomogeneous indices which I don't know from where they originate.

How I implemented this in Mathematica is by doing the following.

packageList["xAct"] = {"xAct`xPerm`","xAct`xTensor`","xAct`xPert`","xAct`Invar`","xAct`xCoba`","xAct`SymManipulator`","xAct`xTras`","xAct`TexAct`","xAct`xPand`"}; System`$PrePrint = xAct`xTensor` ScreenDollarIndices; xAct`xTensor`$DefInfoQ = False; xAct`xTensor`$CovDFormat = "Prefix"; 

Next, we define a three-dimensional manifold $M$ and a metric $\gamma_{ij}$:

DefManifold[M, 3, IndexRange[a, m]]; DefMetric[+1, gamma[-a, -b], cd, SymbolOfCovD -> {"|", "D"}, PrintAs -> "\[Gamma]"]; 

On top of that, we define a conformal metric $\tilde{\gamma}_{ij} =\chi\gamma_{ij}$ which I define with a bar (instead of a tilde).

DefTensor[chi[], M, PrintAs -> "\[Chi]"] DefMetric[+1, gammabar[-a, -b], cdbar, {",", "\!\(\*OverscriptBox[\(D\), \(_\)]\)"}, ConformalTo -> {gamma[-a, -b], chi[]}, PrintAs -> "\!\(\*OverscriptBox[\(\[Gamma]\), \(_\)]\)"] DefTensor[\[Alpha][], M] DefTensor[\[Beta][a], M] DefScalarFunction[p, PrintAs -> "a"]; DefScalarFunction[q, PrintAs -> "b"]; DefTensor[x[], M]; 

As an illustration, we define the unit vector $\xi_{i}$ as follows, together with a rule stating that its length is $1$.

DefTensor[xi[a], M, PrintAs -> "\[Xi]"] xi /: xi[-a_] xi[a_] := 1 

We are now ready to define the matrix

mat1 = Table[Table[0, {10}], {10}]; mat1[[1, 3]] = -2 \[Alpha][] gammabar[-i, c] gammabar[-j, d](*delta[-i,c]delta[-j,d]*); mat1[[1, 8]] = 2 I Symmetrize[gammabar[-h, -i] xi[-j], {-i, -j}] - 2 I/3 gammabar[-i, -j] xi[-h]; mat1[[2, 4]] = 2/3 chi[] \[Alpha][]; mat1[[2, 8]] = -2 I/3 chi[] xi[-h]; mat1[[3, 1]] = 1/2 \[Alpha][] (delta[-a, e] delta[-b, f] - 1/3 gammabar[-a, -b] gammabar[e, f]); mat1[[3, 2]] = -\[Alpha][]/ 2 (xi[-a] xi[-b] - 1/(3 chi[]) gammabar[-a, -b]); mat1[[3, 6]] = I \[Alpha][] chi[] (Symmetrize[gammabar[-g, -a] xi[-b], {-a, -b}] - 1/3 gammabar[-a, -b] xi[-g]); mat1[[3, 7]] = chi[] (xi[-a] xi[-b] - 1/(3 chi[]) gammabar[-a, -b]); mat1[[4, 1]] = 3 q[x[]] \[Alpha][]/(4 (1 + q[x[]])) (1/chi[] xi[e] xi[f] - 1/2 gammabar[e, f]) + 1/2 gammabar[e, f] \[Alpha][] (4 + q[x[]])/(4 (1 + q[x[]])); mat1[[4, 2]] = -2/chi[] \[Alpha][] (4 + q[x[]])/(4 (1 + q[x[]])); mat1[[4, 6]] = I \[Alpha][] chi[] xi[-g]; mat1[[4, 7]] = 1; mat1[[5, 1]] = \[Alpha][]/(2 (1 + q[x[]])) gammabar[e, f]/2 + q[x[]] \[Alpha][]/(2 (1 + q[x[]])) (xi[e] xi[f]/chi[] - gammabar[e, f]/2); mat1[[5, 2]] = -\[Alpha][]/(2 (1 + q[x[]])) 2/chi[]; mat1[[5, 6]] = I \[Alpha][] chi[] xi[-g]/2; mat1[[6, 3]] = I \[Alpha][]/(chi[] (1 + q[x[]])) (-2 chi[] q[x[]] xi[-d] delta[ k, -c]); mat1[[6, 4]] = I \[Alpha][]/(chi[] (1 + q[x[]])) (-4/3 xi[k]); mat1[[6, 5]] = I \[Alpha][]/(chi[] (1 + q[x[]])) 2 xi[k]; mat1[[6, 8]] = -1/chi[] (delta[-h, k] + xi[k] xi[-h]/3); mat1[[7, 4]] = -2 \[Alpha][]/(1 + p[x[]]); mat1[[7, 5]] = 4 \[Alpha][]/(1 + p[x[]]); mat1[[8, 6]] = 3/(4 (1 + p[x[]])) delta[l, -g]; mat1[[8, 7]] = -I p[x[]]/(1 + p[x[]]) \[Alpha][] xi[l]; mat1[[9, 10]] = -\[Alpha][]; mat1[[9, 10]] = -\[Alpha][]; 

I get an error when I apply

Eigenvalues[mat1] 

Hence my question: is it possible to write down this matrix (and if its correct) and is this the correct way to do this?