Matrices with an LU factorisation

The open dense set. Let $\Delta_k(A)$ be the $k\times k$ leading principal minor (top-left $k\times k$ block). The product $p=\Delta_1\cdots\Delta_n$ is a polynomial with $p(I)=1\ne0$, so $\Omega=\{p\ne0\}$ is open and dense. We show every $A\in\Omega$ has an LU factorisation.

Induction on $n$. For $n=1$, $A=(a)=(1)\cdot(a)$. Assume the claim for $n-1$ and write
$$A=\begin{pmatrix}A'&b\\c^T&d\end{pmatrix},$$
with $A'$ of size $n-1$ and all leading minors of $A'$ nonzero (they are $\Delta_1,\dots,\Delta_{n-1}$ of $A$). By induction $A'=L'U'$ with $L'$ unit lower triangular. Since $\det A'=\det U'\ne0$, $U'$ is invertible, and $L'$ is invertible too.

Constructing the factors. Look for
$$L=\begin{pmatrix}L'&0\\v^T&1\end{pmatrix},\qquad U=\begin{pmatrix}U'&u\\0&\delta\end{pmatrix}.$$
Then $LU=\begin{pmatrix}L'U'&L'u\\v^TU'&v^Tu+\delta\end{pmatrix}$. To match $A$ we need
$$L'u=b\ (\Rightarrow u=L'^{-1}b),\qquad v^TU'=c^T\ (\Rightarrow v^T=c^TU'^{-1}),\qquad\delta=d-v^Tu.$$
All three are solvable, so $A=LU$ (and $\delta=\det A/\det A'\ne0$).

Remark. The condition is not necessary: $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has an LU factorisation with $L=I$. It fails for some invertible matrices, like $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ (if it were $LU$, $L=\begin{pmatrix}1&0\\\ell&1\end{pmatrix}$ and $U=\begin{pmatrix}u_{11}&u_{12}\\0&u_{22}\end{pmatrix}$ would give $u_{11}=0$ and then $\ell u_{11}=1$, impossible). This is why pivoting is needed in practice. Lesson. Genericity statements often come from "no pivot vanishes".