Let $A$ be a $n\times n$ matrix, with the principle minor being nonzro, that is, $det\begin{pmatrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{k1}&\cdots&a_{kk} \end{pmatrix}\neq 0$ for $1\leq k\leq n$. Show that there exists a lower triangular matrix $B$ such that $BA$ is upper triangular.
How to do this? What I know and what I can do is as follows: by Gram-Schmidt orthogonal process, there exists an orthogonal matrix $P$ such that $PA$ is upper triangular. How this to be modified?
A weaker hypothesis will suffice . . .
Claim:
If $K$ is a field and $A\in GL_n(K)$, there exists a nonzero lower triangular matrix $B\in M_n(K)$ such that $BA$ is upper triangular.
Proof:
Let $L\subset M_n(K)$ be the set of lower triangular matrices, and let $U\subset M_n(K)$ be the set of upper triangular matrices.
Regarding $M_n(K)$ as a vector space over $K$, and $L,U$ as vector subspaces, we have $$\dim(M_n(K))=n^2$$ and $$\dim(L)=\dim(U)=\frac{n^2+n}{2}$$
Then the set $S=\{BA{\,\mid\,}B\in L\}\;$is a vector subspace of $M_n(K)$, and since $A$ is invertible, we get $$\dim(S)=\dim(L)=\frac{n^2+n}{2}$$
Then we have $$ \dim(S)+\dim(U)=n^2+n > n^2 = \dim(M_n(K)) $$ hence $S\cap U\ne \{0\}$.
The conclusion follows.