A square matrix has an unpivoted unit-lower LU factorisation exactly when all leading principal minors are nonzero

Suppose $A=LU$ as in [L1] and that the diagonal of $U$ is nonzero. For each $k$, the leading block also factors as $A_k=L_k{U}_k$, where $L_k$ is unit lower triangular and $U_k$ is upper triangular with diagonal $u_{00},\dots ,u_{k-1,k-1}$. By [L3] and [L4], ${\mathrm{\Delta }}_k=\det (A_k)=\det (L_k)\det (U_k)=1\cdot {\prod }_{i<k}{u}_{ii}\mathrm{\ne }0.$

For the converse, use induction on $n$. When $n=1$, the condition is ${\mathrm{\Delta }}_1=a_{00}\mathrm{\ne }0$, and $A=(1)(a_{00})$ is already a normalised LU factorisation with nonzero diagonal.

Assume $n>1$ and the converse for size $n-1$. Write $A=\left(\begin{array}{cc}a& r^{\mathsf{T}}\\ c& C\end{array}\right)\mathrm{.}$ Then $a={\mathrm{\Delta }}_1\mathrm{\ne }0$. Set $S:=C-c{a}^{-1}{r}^{\mathsf{T}}$. For each $1\le j<n$, apply [L2] to the leading $(j+1)\times (j+1)$ block of $A$: its determinant is $a\det (S_j)$, where $S_j$ is the leading $j\times j$ block of $S$. Hence $\det (S_j)={\mathrm{\Delta }}_{j+1}\mathrm{/}a\mathrm{\ne }0.$ By the induction hypothesis, $S$ has a normalised LU factorisation $S=\tilde{L}\tilde{U}$ with nonzero diagonal.

Step 1.3 and [L2] give $A=\left(\begin{array}{cc}1& 0\\ c{a}^{-1}& \tilde{L}\end{array}\right)\left(\begin{array}{cc}a& r^{\mathsf{T}}\\ 0& \tilde{U}\end{array}\right),$ which is a normalised LU factorisation of $A$. Its upper diagonal is $a$ followed by the diagonal of $\tilde{U}$, so it is nonzero.

Step 1.1 proves the forward implication, and steps 1.2-2.1 prove the reverse implication for every $n\ge 1$. The final sentence is the case $k=n$.