Triangular matrices proof Let $A$, $B$ be $n\times n$ lower triangular matrices.
Prove that $AB$ is also a lower triangular matrix.
How do I prove this for every $n$? 
 A: Induction and block partitioning of matrices is a very useful tool for proving statements like this.
The statement is trivial for $n=1$.
Assume that the products of $n\times n$ ($n\geq 1$) lower triangular matrices are lower triangular and let $A$ and $B$ be $(n+1)\times (n+1)$. Consider the paritioning
$$
A:=\pmatrix{\alpha_{11} & 0 \\ a_{21} & A_{22}}, \quad
B:=\pmatrix{\beta_{11} & 0 \\ b_{21} & B_{22}},
$$
where $A_{22}$ and $B_{22}$ are $n\times n$. We have
$$
C:=AB=
\pmatrix{\alpha_{11} & 0 \\ a_{21} & A_{22}}
\pmatrix{\beta_{11}  & 0 \\ b_{21} & B_{22}}
=
\pmatrix{\alpha_{11}\beta_{11} & 0\\\beta_{11} a_{21}+A_{22}b_{21}&A_{22}B_{22}}.
$$
Hence $C$ is lower triangular if and only if $A_{22}B_{22}$ is lower triangular. But $A_{22}$ and $B_{22}$ are $n\times n$ lower triangular and by the induction assumption, their product is lower triangular. Hence $C$ is lower triangular.
A: Hint: Suppose that $A,B$ are $n\times n$ matrices. Then if $i,j\in\{1,...,n\},$ we have that the $i^\text{th}$ row, $j^\text{th}$ column entry of $AB$ is $$\sum_{k=1}^na_{ik}b_{kj}.$$ You must show that this is $0$ when $i<j$ (note that this differs from what you said in the comments above!) so long as $A,B$ are lower triangular.
A: I'd rather prove this using transformations associated to both matrices A and B. They are both assumed to be nxn square matrices.
Let u be the tranformation associated to A in the canonical basis $(e_1,..,e_n)$. Let v be the same for B.
I'll define: $ I_j = Vect(e_j,..,e_n). $ We obviously have : $I_{j+1} \subset I_j$
What does :"A and B lower triangular matrices " mean?
It means that: for every $k\in (1..n):  u(e_k)\in I_k; v(e_k)\in I_k$. This is an equivalency of definition. The zero above the diagonal mean that $u(e_j)$ does not have a part expressed in terms of $e_i$, with $i\in (1..j-1)$.
This implies that: $ u(I_k) \subset I_k; v(I_k) \subset I_k $
Now if I call w the tranformation associated to C=AB, we have: w=u o v (composition of tranformations)
We only have to prove that : for every  $k\in (1..n):  w(e_k)\in I_k$:
$ w(e_k) = u(v(e_k)) $ . But $v(e_k)\in I_k= Vect(e_k,..,e_n)$, so as it's been stated before: $u(I_k) \subset I_k \implies w(e_k) = u(v(e_k))\in I_k$
So w is a transformation associated to a lower triangular matrix
A: Hint: Write it down in a general form:
$$ \left( \begin{array}{ccc}
a_{11} & \cdots & a_{1n} \\
\vdots & \cdots & \vdots \\
a_{n1} & \cdots & a_{nn} \end{array} \right) \cdot \left( \begin{array}{ccc}
b_{11} & \cdots & b_{1n} \\
\vdots & \cdots & \vdots \\
b_{n1} & \cdots & b_{nn} \end{array} \right) $$
Now for lower triangular matrices you have that the upper parts of the matrices are all zeros.
Now just calculate using the definition of matrix multiplication. It is sufficient to show that again all upper triangular part and the diagonal is zero ...
