Strict complementary slackness for semidefinite programs with strong duality By a theorem of Goldman and Tucker it is known that if a linear program (LP) has a finite valued optimal solution, then there is an optimal primal/dual pair $(x,z)$ satisfying not only complementary slackness (i.e. $x_iz_i=0$), but also strict complementary slackness (i.e. exactly one of $x_i$ and $z_i$ vanishs). 
It is known that this property does not carry over to semidefinite programs (SDP). This means, if $(X,Z)$ is an optimal primal/dual pair of an SDP, where $X,Z\in\Bbb R^{n\times n}$ are symmetric positive semi-definite matrices, then we have complementarity (i.e. $XZ=0$), but in general no strict complementarity $\def\rank{\mathrm{rank}}\rank(X)+\rank(Z)=n$.
Now, I am especially interested in SDPs with finite valued optimal solutions and zero duality gap. Is there anything more we can say about strict complementary slackness in the case of strong duality? I actually have never seen an example where it fails, i.e. an example with above properties where all optimal primal/dual pairs $(X,Z)$ lack strict complementarity.
Most of my knowledge about strict complementarity comes from [1] where it is shown that it is a generic property (holds for almost all SDPs in a precise sense). Besides this, strict complementarity was mostly assumed to prove other properties.
[1] F. Alizadeh, J.A. Haeberly, M.L. Overton: Complementarity and nondegeneracy in semidefintie programming
 A: Studying certain semidefinite programs arising in spectral graph theory, I discovered a semi-definite primal/dual pair satisfying strong duality but not strict complementarity.
Let $E=\{12,23,34\}$ (the edge set of the path graph $P_3$) and $E^{ij}:=(\mathbf e_i-\mathbf e_j)^\top(\mathbf e_i-\mathbf e_j)$. We use the inner product $
\def\<{\langle}\def\>{\rangle}\<X,Y\>:=\mathrm{tr}(XY)$ on the space of symmetric matrices $\mathbf S^n$.
$$
\llap{\mathrm{(P)}\qquad}
\boxed{\begin{array}{rlr}
p^*=\max & \sum_{ij\in E} w_{ij} \\
\mathrm{s.t.} & I-\sum_{ij\in E} w_{ij}E^{ij} \succeq0 \\
              & w_{ij}\ge 0,\quad ij\in E
\end{array}}$$
$$
\llap{\mathrm{(D)}\qquad}
\boxed{\begin{array}{rlr}
d^*=\min & \mathrm{tr}(X) \\
\mathrm{s.t.} & \<X,E^{ij}\>\ge 1,\quad ij\in E \\
              & X\succeq 0
\end{array}}$$
The common optimal value is $p^*=d^*=1$, obtained for
$$w_{12}=w_{34}=\frac12,\quad w_{23}=0,$$
$$X=\frac14\begin{pmatrix}
\phantom+1 & -1 & \phantom+1 & -1 \\
-1 & \phantom+1 & -1 & \phantom+1 \\
\phantom+1 & -1 & \phantom+1 & -1 \\
-1 & \phantom+1 & -1 & \phantom+1
\end{pmatrix} 
= \frac14 (\phantom+1,-1,\phantom+1,-1)^\top(\phantom+1,-1,\phantom+1,-1).$$
Both are the unique solutions of their respective problem (this can easily be seen when interpreting $\mathrm{(D)}$ as an embedding problem, but I will leave this out here). As seen, the dual positive semi-definite matrix $X$ has rank $1$. However, the corresponding matrix
$$Z:=I-\sum_{ij\in E} w_{ij}E^{ij} = 
\frac12 \begin{pmatrix} 
1 & 1 &   &   \\
1 & 1 &   &   \\ 
  &   & 1 & 1 \\
  &   & 1 & 1
\end{pmatrix}$$
is of rank $2$. Hence $\mathrm{rank}(X)+\mathrm{rank}(Z)=3<4$ and strict complementarity is not satisfied.
A: Take the SDP program 
\begin{align*}
\min~&  x_3\\
s.t.~&
X = \begin{bmatrix}
        x_1   &        x_2 &    0      &     0    \\
        x_2   &        0   &    0      &     0    \\
        0     &        0   &    x_2    &     0    \\
        0     &        0   &    0      &   x_3-2  \\
\end{bmatrix}\succeq \mathbf{0}\\
&x_1,~x_2,~x_3\in\mathbf{R}
\end{align*}
Its dual can be written as follows (see below why).
\begin{align*}
\max~&  2z_{44}\\
s.t.~& Z=
\begin{bmatrix}
        0     &        0   &    0      &     0    \\
        0     &     a^2+\Delta^2&    0      &     a    \\
        0     &        0   &    0      &     0    \\
        0     &        a   &    0      &     1  \\
\end{bmatrix}\\
&Z\succeq 0
\end{align*}
Now notice that any feasible solution $\mathbf{x}$ of the primal satisfies
$x_2=0$ because of the zero situated at position $(2,2)$; recall that any SDP matrix $A$ such that $A_{ii}=0$ has only zeros on row and column $i$.
Now check that $X$ and $Z$ share the eigenvector 
$
\left[\begin{smallmatrix}
        0     &
        0        &
        1        &
        0        \\
\end{smallmatrix} \right]^\top$ with eigenvalue 0, which actually answers the question. No linear combination of the rows of $X$ and $Z$ that can be equal to this eigenvector, i.e., the rows of $X$ and $Z$ do not cover the whole space $\implies rank(X)+rank(Z)<n$.
The only detail that remains to be filled is to show that the above expression of the dual is correct. Any feasible $Z$ satisfies $z_{11}=0$ because the coefficient 
of $x_1$ is zero in the primal objective function. This forces row 1 and column 1 of $Z$ to have only zeros. The dual
constraint corresponding to $x_2$ is $2z_{12}+z_{33}=0$; since $z_{12}=0$, we
have $z_{13}=0$. The dual constraint corresponding to $x_3$ imposes $z_4=1$.
There is no constraint on $z_{24}=a$; $z_{22}$ needs to be greater than or equal
to $a^2$ so as to have a non-negative principal minor corresponding to
rows/columns 2 and 4; we can write
$z_{22}=a^2+\Delta^2$. Finally, both programs have objective value 2.
