Factoring a matrix as the product of block triangular and diagonal matrices. How can I check that the matrix 
$$K = \left[\begin{array}{c|cc} 1 & 0^{\mathrm T}_m & m \mathbf u^{\mathrm T} \\ \hline  0_m & I_m & I_m \\ m \mathbf u & I_m & O_m \end{array}\right] = \begin{bmatrix}
    1       & \textbf{w}^T \\
    \textbf{w}      & B \\
\end{bmatrix}$$
can be factored as the following product of block triangular and diagonal matrices:
\begin{equation} K = 
\begin{bmatrix} 
    1 & \textbf{w}^{T}B^{-1} \\
    \textbf{0} & I_{2m}\\
\end{bmatrix} 
\begin{bmatrix}
    s & \textbf{0}^T\\
    \textbf{0}       & B\\
\end{bmatrix}
\begin{bmatrix}
1 & \textbf{0}^{T}\\
B^{-1}\textbf{w} & I_{2m}
\end{bmatrix}
\end{equation}
where $$B = \begin{bmatrix} I_m & I_m \\ I_m & O_m \end{bmatrix} = \begin{bmatrix} 1 I_m & 1 I_m \\ 1 I_m & 0 I_m \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \otimes I_m.$$ and 
$$\mathbf w = \begin{bmatrix} 0_m \\ m \mathbf u \end{bmatrix} = \begin{bmatrix} 0 \mathbf u \\ m \mathbf u \end{bmatrix} = \begin{bmatrix} 0 \\ m \end{bmatrix} \otimes \mathbf u$$ where $$\mathbf u = \begin{bmatrix} 
1 \\
\vdots \\
1 
\end{bmatrix} \in \mathbb R^m$$ and I need to specify $s$. I think i have calculated the inverse of the Kronecker product $B$ correctly as $$B^{-1} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \otimes I_m =  \begin{bmatrix} 0I_{m} & 1I_{m} \\ 1 I_{m} & -1I_{m} \end{bmatrix}$$ but I don't know where to go from here. Any help would be great! 
 A: It works. To multiply
$\begin{equation} K = 
\begin{bmatrix} 
    1 & \textbf{w}^{T}B^{-1} \\
    \textbf{0} & I_{2m}\\
\end{bmatrix} 
\begin{bmatrix}
    s & \textbf{0}^T\\
    \textbf{0}       & B\\
\end{bmatrix}
\begin{bmatrix}
1 & \textbf{0}^{T}\\
B^{-1}\textbf{w} & I_{2m}
\end{bmatrix}
\end{equation},$
start, e.g., by multiplying the last two matrices. Perform block-wise multiplication like regular matrix multiplication. You get:
$\begin{align} K &= 
\begin{bmatrix} 
    1 & \textbf{w}^{T}B^{-1} \\
    \textbf{0} & I_{2m}\\
\end{bmatrix} 
\begin{bmatrix}
    s + \textbf{0}^T B^{-1} \textbf{w} & s \textbf{0}^T + \textbf{0}^T I\\
    \textbf{0} 1 + B B^{-1} w & \textbf{0}\textbf{0}^T + B I \\
\end{bmatrix} \\
& = \begin{bmatrix} 
    1 & \textbf{w}^{T}B^{-1} \\
    \textbf{0} & I_{2m}\\
\end{bmatrix} 
\begin{bmatrix}
    s  &  \textbf{0}^T \\
    w &   B  \\
\end{bmatrix}
\end{align}.$
Keep going the same way:
$\begin{align}
& K &= \begin{bmatrix} 
    s + \textbf{w}^{T}B^{-1}\textbf{w} & 1\textbf{0}^T + \textbf{w}^T B^{-1} B \\
    \textbf{0} s + I \textbf{w} & \textbf{0}\textbf{0}^T + B\\
\end{bmatrix} = 
\begin{bmatrix}
    s + \textbf{w}^{T}B^{-1}\textbf{w} &  \textbf{w}^T \\
    \textbf{w} &   B  \\
\end{bmatrix}
\end{align}.$
This is equal to your desired $K$ if $s + \textbf{w}^{T}B^{-1}\textbf{w} =1$. Let us evaluate $\textbf{w}^{T}B^{-1}\textbf{w}$:
$\begin{align}
\textbf{w}^{T}B^{-1}\textbf{w}
&= \left( \begin{bmatrix}0\\m\end{bmatrix} \otimes \textbf{u}\right)^T
\left( \begin{bmatrix}0 & 1\\1&-1\end{bmatrix} \otimes I_m\right)
\left( \begin{bmatrix}0\\m\end{bmatrix} \otimes \textbf{u}\right)\\
 &= \left( \begin{bmatrix}0, m\end{bmatrix} \cdot \begin{bmatrix}0 & 1\\1&-1\end{bmatrix}
\cdot \begin{bmatrix}0\\m\end{bmatrix} \right) \otimes \left(
\textbf{u}^T I_m \textbf{u}\right) \\
& = [m, -m] \cdot \begin{bmatrix}0\\m\end{bmatrix}\cdot m = -m^2 \cdot m = -m^3.
\end{align},$
where we used the fact that $(A \otimes B) \cdot (C \otimes D) = (AC \otimes BD)$ if dimensions comply.
Therefore, your desired $s$ must satisfy $s -m^3=1$, i.e., $s = 1+m^3$.
