Singular interval condition for Multi input in Optimal control
1st interpretation of singular interval
&quot;[A singular interval is] where Pontryagin's principle provides no information regarding u in terms of x, p and t.&quot;
You have a minimum-time optimal control which I assume is stated as:
$$ 
\min_{\|u\|_\infty \leq 1} t_f \\
\text{s.t.} \quad \dot{x}(t) = Ax(t) + Bu(t)\\
 x(0) = 0, \quad
 x(t_f) = x_t\\
 0\leq t_f.
$$
And I assume that the pre-Hamiltonian is
$$ H(t,x,p,u) = -p^\top (Ax+Bu),$$
and that you have a costate \$p^*\$ that is not identically zero and evolves as:
$$ \dot{p}^*(t) = -A^\top p^*(t).$$
Your definition seems to mean that in a singular interval any feasible \$u\$ will be the optimal control with the same value for \$H(t,x^*,p^*,u^*)=H(t,x^*,p^*,u)\$, therefore any feasible \$u\$ will have:
$$ 0 = {p^*}^\top B u $$
And you can show that if any \${p^*}^\top B u \$ is nonzero then you would have some information on \$u\$, i.e. the optimal \$u\$ would be at the boundary of \$ \{u : \|u\|_\infty \leq 1\}\$ and in certain direction(s). So:
$$ 0 = {p^*}^\top B u, \, \forall u  \Leftrightarrow {p^*}^\top B =0,$$
And since \$p^*\$  is differentiable we have:
$$ \left(\frac{\partial}{\partial t} \right)^k {p^*}^\top B = {p^*}^\top (-A)^k B=0, $$
which gives us that:
$$ {p^*}^\top \left[ B, AB, A^2B \dots A^nB \right] = 0$$
meaning that the controlability matrix is rank deficient. Somehow I got a stronger result than what the problem asks, there is probably something wrong with my proof and it is probably that definition of singular control, where Pontryagin's  principle provides no information regarding u in terms of x, p and t.
2nd interpretation of singular interval
I decide to interpret that a singular interval is one where Pontryagin's principle does not provide a unique \$u\$ in terms of \$x\$, \$p\$ and \$t\$. That would meant that we can have two (or more) optimal values for \$u\$, let them be \$\tilde{u}, \hat{u}\$. Since we have the constraint \$\|u\|_\infty \leq 1\$ and the cost function is linear we have that the minima of the cost will be achieved at the vertices of the polyhedra.
And
$$ \tilde{u}, \hat{u} \in \{-1,1\}^m. $$
And we can even construct such optima, to determine the components \$\hat{u}_i\$ or \$\tilde{u}\$ we do
$$
 \nabla_u (-p^\top B u) = -B^\top p,\\
 \hat{u}_i = \begin{cases} -1,\quad (-B^\top p)_{i,:} &gt; 0 \\ 1,\quad (-B^\top p)_{i,:} &lt; 0 \\ ?,\quad (-B^\top p)_{i,:} = 0 \end{cases}
$$
where \$ (A)_{i,:}\$ means the i-th row of A and \$ (A)_{:,i}\$ means the i-th column.
Since we have at least two different optima \$\tilde{u}\$ and \$\hat{u}\$ they can only be different in components \$i\$ with \$(-B^\top p)_{i,:} = 0\$, and there is at least one such \$i\$. Then you have
$$ (-B^\top p)_{i,:} = 0 = -(B^\top)_{i,:} p = -p^\top (B)_{:,i} $$
Denoting the i-th column of \$B\$ as \$(B)_{:,i}=b_i\$ we can clearly see that
$$ -p^\top b_i = 0 $$
and due to \$p\$ being differentiable we get that:
$$ -p^\top (-A)^k b_i = 0 $$
and
$$ p^\top [ b_i, A b_i, A^2 b_i, \dots ] = 0. $$
Which means that \$E_i\$ (as defined in your question) will be singular.