TITLE: Proving the inverse (if any) of a lower triangular matrix is lower triangular
QUESTION [22 upvotes]: The inverse of a non-singular lower triangular matrix is lower triangular.
Construct a proof of this fact as follows. Suppose that $L$ is a non-singular lower triangular matrix. If $b \in \mathbb{R^n}$ is such that $b_i = 0$ for $i = 1, . . . , k \leq n$, and $y$ solves $Ly = b$, then $y_i = 0$ for
$i = 1, . . . , k \leq n$. (Hint: partition $L$ by the first $k$ rows and columns.)

Can someone tell me what exactly we are showing here and why it will prove that the inverse of any non-singular lower triangular matrix is lower triangular?

REPLY [2 votes]: I was thinking about this same question and have an explanation from an informal perspective:
With invertible matrix $A$ and $$LA = B$$ 
We know that the first row of $B$ is a multiple of the first row of $A$, and the second row of $B$ is a linear combination of the first two rows of $A$, ..., the $k$th row of $B$ is a linear combination of the first $k$ rows of $A$,...
It follows that for any $k$, the first $k$ rows of $A$ and the first $k$ rows of $B$ span the same subspace. Therefore the $k$th row of $A$ is in the subspace spanned by the first $k$ rows of $B$. Furthermore, the $k$th row of $A$ cannot be in the subspace spanned by the first $k-1$ rows of $B$. Otherwise it is in the subspace spanned by the first $k-1$ rows of $A$, which contradicts the assumption that $A$ is invertible (rows are linearly independent). Because for any $k$ the $k$th row of $A$ is a linear combination of the first $k$ rows of $B$, for $$L^{-1}B=A$$ $L^{-1}$ must be lower triangular.