Product of lower triangular matrices We have the $n\times n$-matrices $A=I-y_i\cdot e_i^T$ and $B=I+y_i\cdot e_i^T$, where $e_i$ is the $i$-th unit vector and  at the vector $y_i$ we have that the elements are $y_{k,i}$ where for $k<i$ it holds that $y_{k,i}=0$ and $y_{i,i}=1$ and for $k>i$ we have that $y_{k,i}\in \mathbb{R}$. 
(So, $A$ and $B$ are lower triangular with $1$ at the diagonal.)
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I want to calculate their product.  
We have the following: 
\begin{align*}AB&=(I-y_i\cdot e_i^T)\cdot (I+y_i\cdot e_i^T)=I+y_i\cdot e_i^T-y_i\cdot e_i^T-(y_i\cdot e_i^T)\cdot (y_i\cdot e_i^T)\\ & =I-(y_i\cdot e_i^T)\cdot (y_i\cdot e_i^T)\end{align*} 
right? 
Can we calculate the product $(y_i\cdot e_i^T)\cdot (y_i\cdot e_i^T)$ ? 
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EDIT: 

What I actually want to show is that the product of these two matrices is equal to the identity matrix. Is the way I wrote the matrices wrong for this purpose? 
 A: What you have written down in your edit does indeed equal $I$, but you have done some mistake calculating $A$ and $B$, since, the way you have defined them, they actually are
$$
A =
\begin{pmatrix}
1 & \dots & 0 & 0 & 0 & \dots & 0\\
\vdots &  & \vdots &  & \vdots & & \vdots \\
0 & \dots & 1 & 0 & 0 & \dots & 0\\
0 & \dots & 0 & 0 & 0 &\dots\ & 0 \\
0 & \dots & 0 & -y_{i+1,i} & 1 & \dots & 0\\
\vdots &  & \vdots & \vdots & \vdots &  & \vdots\\
0 & \dots & 0 & -y_{n,i} & 0 & \dots & 1
\end{pmatrix}
~~~~~~~~~~~~B = 
\begin{pmatrix}
1 & \dots & 0 & 0 & 0 & \dots & 0\\
\vdots &  & \vdots &  & \vdots & & \vdots \\
0 & \dots & 1 & 0 & 0 & \dots & 0\\
0 & \dots & 0 & 2 & 0 &\dots\ & 0 \\
0 & \dots & 0 & y_{i+1,i} & 1 & \dots & 0\\
\vdots &  & \vdots & \vdots & \vdots &  & \vdots\\
0 & \dots & 0 & y_{n,i} & 0 & \dots & 1
\end{pmatrix}
$$
Which does not yield $I$ when multiplied.
The most you can simplify $(y_i\cdot e_i^T)\cdot(y_i\cdot e_i^T)$ is propably the following
$$
(y_i\cdot e_i^T)\cdot(y_i\cdot e_i^T) 
= 
y_i\cdot \underbrace{(e_i^T \cdot y_i )}_{=1} \cdot e_i^T
=
y_i \cdot e_i^T 
$$
So you finally get
$$ AB = A $$
(And even $BA = A$)
