& \vdots\\
  0 & \cdots & {\Lambda}_{{O}}^{t_1}\\ 
   \hline
   \vdots & \ddots & \vdots\\
   \hline
  {\Lambda}_1^{t_{{L}}} & \cdots & 0\\
   \vdots & \ddots & \vdots\\
  0 & \cdots & {\Lambda}_{{O}}^{t_{L}}   
 \end{bmatrix}}_{{\widehat{\mathbf{A}}}\in{\mathbb{R}}^{{M}{L}\times{M}}}$$ Recall that a matrix’s rank is preserved under a product with an invertible matrix. Design a matrix $U\in{\mathbb{R}}^{{N}\times{N}}$ s.t. ${\widetilde{{\ensuremath{K}}}}:= U{\ensuremath{K}}$ is a matrix with one row vector of nonzeros, and all of the remaining rows as zeros. Then $\operatorname{rank}({\widehat{\mathbf{K}}}{\widehat{\mathbf{A}}}) = \operatorname{rank}(U{\widehat{\mathbf{K}}}{\widehat{\mathbf{A}}})$. Therefore, we have that $${\widetilde{{\ensuremath{K}}}}A^{t_j} = 
 \begin{bmatrix}
  {\widetilde{{\ensuremath{K}}}}_{(1)}\\
  0\\
  \vdots\\
  0
 \end{bmatrix}
 A^{t_j}\\
 = 
 \begin{bmatrix}
  k_{11}{\lambda}_1^{t_j} & \binom{t_j}{1}{\lambda}_1^{t_j-1} + k_{12}{\lambda}_1^{t_j}  & \cdots & k_{1{M}}{\lambda}_{{O}}^{t_j}\\
  0 & 0 & \cdots & 0\\
  \vdots & \vdots & \ddots & 0\\
  0 & 0 & \cdots & 0
 \end{bmatrix}$$ Therefore, following some more elementary row operations encoded by $V\in{\mathbb{R}}^{{M}{L}\times{M}{L}}$, we get that $$V \begin{bmatrix}
    {\widetilde{{\ensuremath{K}}}}& \cdots & {\widetilde{{\ensuremath{K}}}}\end{bmatrix}
   \begin{bmatrix}
    A^{t_1}\\
    \vdots\\
    A^{t_{{L}}}
   \end{bmatrix}
   = 
   \begin{bmatrix}
    \tilde{k}_{11}{\lambda}_1^{t_1} & \cdots & \tilde{k}_{1{M}}{\lambda}_{{O}}^{t_1}\\
    \tilde{k}_{11}{\lambda}_1^{t_2} & \cdots & \tilde{k}_{1{M}}{\lambda}_{{O}}^{t_2}\\
    \vdots & \ddots & 0\\
    \tilde{k}_{11}{\lambda}_1^{t_{{L}}} & \cdots 
    & \tilde{k}_{1{M}}{\lambda}_{{O}}^{t_{{L}}}\\
    \mathbf{0} & \cdots & \mathbf{0}
   \end{bmatrix}\\
   = 
   \begin{bmatrix}
    \boldsymbol{\Phi}\\
    \widehat{\mathbf{0}}
   \end{bmatrix}.$$ If the individual entries $\tilde{k}_{1i}$ are nonzero, and the Jordan block diagonals have nonzero eigenvalues, the columns of $\boldsymbol\Phi$ become linearly independent. Therefore, if ${L}\geq 