[proofplan] We prove existence of the LU factorisation by induction on the matrix dimension, using the block structure of $A$ and the non-singularity of leading submatrices. Uniqueness follows from the fact that a matrix simultaneously unit lower triangular and upper triangular must be the identity. [/proofplan] [step:Prove existence by induction using the block partition] For $n = 1$: $A = (a_{11}) = (1)(a_{11}) = LU$. For the inductive step, partition $A = \begin{pmatrix} A_{n-1} & c \\ r^\top & a_{nn} \end{pmatrix}$. Since $\det(A_k) \neq 0$ for $k = 1, \ldots, n-1$, the inductive hypothesis gives $A_{n-1} = L_{n-1}U_{n-1}$ uniquely. Moreover $U_{n-1}$ is non-singular ($\det(U_{n-1}) = \det(A_{n-1}) \neq 0$). Seek $L = \begin{pmatrix} L_{n-1} & \mathbf{0} \\ \ell^\top & 1 \end{pmatrix}$ and $U = \begin{pmatrix} U_{n-1} & u \\ \mathbf{0}^\top & u_{nn} \end{pmatrix}$. Matching $LU = A$: (i) $L_{n-1}u = c$ gives $u = L_{n-1}^{-1}c$ (unique). (ii) $\ell^\top U_{n-1} = r^\top$ gives $\ell^\top = r^\top U_{n-1}^{-1}$ (unique). (iii) $u_{nn} = a_{nn} - \ell^\top u$ (determined uniquely). [/step] [step:Prove uniqueness] Suppose $A = L_1 U_1 = L_2 U_2$. Then $L_2^{-1}L_1 = U_2 U_1^{-1}$. The left side is unit lower triangular; the right side is upper triangular. A matrix that is simultaneously unit lower triangular and upper triangular must be $I$. Therefore $L_1 = L_2$ and $U_1 = U_2$. [/step]