type I: Under a Lorentz transformation we can set

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^{\mu\nu} \sim_{\rm L} \left[\begin{array}{@{}c|ccc@{}} \rho&0 &0 &0\nonumber \\ \hline 0 &p_1& 0 &0\nonumber \\ 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]

\qquad T^\mu{}_\nu \sim_{\rm L}\left[\begin{array}{@{}c|ccc@{}} -\rho&0 &0 &0\nonumber \\ \hline 0 &p_1& 0 &0\nonumber \\ 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]. \nonumber \end{align} \tag{ 3 } As in this case the stress-energy tensor is fully diagonalizable, one obtains the same matrix by expressing this tensor in the orthonormal basis formed by its eigenvectors

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^\mu{}_\nu \sim\left[\begin{array}{@{}c|ccc@{}} -\rho&0 &0 &0\nonumber \\ \hline 0 &p_1& 0 &0\nonumber \\ 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]. \nonumber \end{align} \tag{ 4 } The eigenvalues are \{-\rho, p_1, p_2, p_3\}. This is as simple as type I gets. In (3+1) dimensions type I is invariantly characterized by the existence of a unique timelike eigenvector, implying the existence of three spacelike eigenvectors. (In Euclidean signature, since \newcommand{\e}{{\rm e}} \eta^{ab} \to \delta^{ab}, all stress-energy tensors are type I.)

type II: Under a Lorentz transformation we can set

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^{\mu\nu} \sim_{\rm L} \left[\begin{array}{@{}cc|cc@{}} \mu+f&f &0 &0\nonumber \\ f &-\mu+f& 0 &0\nonumber \\ \hline 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]

\quad T^\mu{}_\nu \sim_{\rm L} \left[\begin{array}{@{}cc|cc@{}} -\mu-f&f &0 &0\nonumber \\ -f &-\mu+f& 0 &0\nonumber \\ \hline 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]

\nonumber \end{align} \tag{ 5 } while under a generic non-singular similarity transformation we can set

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^\mu{}_\nu \sim\left[\begin{array}{@{}cc|cc@{}} -\mu&1 &0 &0\nonumber \\ 0 &-\mu& 0 &0\nonumber \\ \hline 0&0& p_2 & 0\nonumber \\0&0&0&p_3\nonumber \end{array}\right]. \nonumber \end{align} \tag{ 6 } The eigenvalues are \{-\mu, -\mu, p_2, p_3\}. The non-trivial structure of the Jordan form is due to the null eigenvector associated to the double eigenvalue. That is, it is not expressed in a basis with a timelike eigenvector, though there are two spacelike eigenvectors associated to the pi eigenvalues. As a matrix, type II is defective, there are only three eigenvectors, not four. Now from type II subtract out as much of type I as possible

and call what is left type II0. (Make the eigenvalues as simple as possible

but do not disturb the eigenvectors.)Consider this:

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (T^{\mu\nu})_{\rm II_0} \sim_{\rm L} \left[\begin{array}{@{}cc|cc@{}} f&f &0 &0\nonumber \\ f &f& 0 &0\nonumber \\ \hline 0&0&0 & 0\nonumber \\0&0&0&0\nonumber \end{array}\right]

\quad (T^\mu{}_\nu)_{\rm II_0}\sim_{\rm L} \left[\begin{array}{@{}cc|cc@{}} -f&f &0 &0\nonumber \\ -f &f& 0 &0\nonumber \\ \hline 0&0&0 & 0\nonumber \\0&0&0&0\nonumber \end{array}\right]

\nonumber \end{align} \tag{ 7 } and

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (T^\mu{}_\nu)_{\rm II_0} \sim\left[\begin{array}{@{}cc|cc@{}} 0&1 &0 &0\nonumber \\ 0 &0& 0 &0\nonumber \\ \hline 0&0& 0& 0\nonumber \\0&0&0&0\nonumber \end{array}\right]. \nonumber \end{align} \tag{ 8 } The eigenvalues are now as simple as possible, \{0, 0, 0, 0\}, and there is still one null and two spacelike eigenvectors. Observe that type II0 can be invariantly characterized by the equation [(T^a{}_b)_{\rm II_0}]^2 = 0. (So it is nilpotent of order 2.) We can also write \newcommand{\e}{{\rm e}} (T^{ab})_{\rm II_0} = f \ell^a \ell^b where \newcommand{\e}{{\rm e}} \ell is a null vector with unit time component. On the other hand, type II0 tensors have a generalized timelike eigenvector, such that \newcommand{\e}{{\rm e}} [(T^a{}_b)_{\rm II_0}]t^b=f\ell^a, implying [(T^a{}_b)_{\rm II_0}]^2 t^b = 0. The minimum dimensionality for a type II0 stress-energy tensor is (1+1)

type III: Under a Lorentz transformation we can set

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^{\mu\nu} \sim_{\rm L} \left[\begin{array}{@{}ccc|c@{}} \rho&f &0&0\nonumber \\ f &-\rho& f &0\nonumber \\ 0&f& -\rho & 0\nonumber \\ \hline 0&0&0&p_3\nonumber \end{array}\right]

\quad T^\mu{}_\nu \sim_{\rm L} \left[\begin{array}{@{}ccc|c@{}} -\rho&f &0&0\nonumber \\ -f &-\rho& f &0\nonumber \\ 0&f& -\rho & 0\nonumber \\ \hline 0&0&0&p_3\nonumber \end{array}\right]

\nonumber \end{align} \tag{ 9 } where the parameter f is unnecessarily set to 1 in reference [, 1: #cqgaac147bib001]. The Jordan normal form of this tensor is

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T^\mu{}_\nu \sim\left[\begin{array}{@{}ccc|c@{}} -\rho&1 &0 &0\nonumber \\ 0 &-\rho& 1 &0\nonumber \\ 0&0& -\rho & 0\nonumber \\ \hline 0&0&0&p_3\nonumber \end{array}\right]. \nonumber \end{align} \tag{ 10 } The eigenvalues are \{-\rho, -\rho, -\rho, p_3\}, and there is a single null eigenvector associated to the triple eigenvalue, plus a single spacelike eigenvector associated to the eigenvalue p3. As a matrix type III is defective, there are only two eigenvectors, not four.Now from type III subtract out as much of type I as possible

call what is left type III0. (Make the eigenvalues as simple as possible

but do not disturb the eigenvectors.) Consider this:

\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (T^{\mu\nu})_{\rm III_0} \sim_{\rm L} \left[\begin{array}{@{}ccc|c@{}} 0&f &0&0\nonumber \\ f &0& f &0\nonumber \\ 0&f&0& 0\nonumber \\ \hline 0&0&0&0\nonumber \end{array}\right]

\quad (T^\mu{}_\nu)_{\rm III_0} \sim_{\rm L} \left[\begin{array}{@{}ccc|c@{}} 0&f &0&0\nonumber \\ -f &0& f &0\nonumber \\ 0&f&0& 0\nonumber \\ \hline 0&0&0&0\nonumber \end{array}\right]

\nonumber \end{align} \tag{ 11 } with Jordan form