Proving by induction that $L = I + \sum_{ k=1}^{n-1} u ^{(k)} e ^T _k $ Let $L = \prod_{k=1}^{n-1}(I + u^{(k)}e^T_k) $ where $u^{(k)} 
(i) = 0$ for $i = 1 : k$.
Prove that $L = I + \sum_{k=1}^{n-1}u^{(k)} e^T_k$
by induction. (where $L$ is the lower triangular matrix of the $LU$ decomposition of a matrix $A$).
My attempt:
Base case: $n=2$
$\prod_{k=1}^{1}(I + u^{(k)}e^T_k) = I + \sum_{k=1}^{1}u^{(k)} e^T_k$
$I + u^{(1)}e^T_1 = I + u^{(1)}e^T_1$. True.
Now assume true for $n=j$, let's show it's true for $n=j+1$.
$\prod_{k=1}^{j+1-1}(I + u^{(k)}e^T_k) = I + \sum_{k=1}^{j+1-1}u^{(k)} e^T_k$
Then, $\prod_{k=1}^{j}(I + u^{(k)}e^T_k) = I + \sum_{k=1}^{j}u^{(k)} e^T_k$
Which is also true, because it is the assumption we made for $n=j$, which is true.
Is my attempt correct?
 A: I cannot see where do you actually prove that the validity of the identity for $n=j$ implies its validity for $n=j+1$.
You assume that the identity works for $n=j$, that is,
$$\tag{1}
\prod_{k=1}^{j-1}(I+u^{(k)}e_k^T)=I+\sum_{k=1}^{j-1}u^{(k)}e_k^T.
$$
Now you have to prove (mathematically, not by a political statement) that this implies that
$$\tag{2}
\prod_{k=1}^{j}(I+u^{(k)}e_k^T)=I+\sum_{k=1}^{j}u^{(k)}e_k^T,
$$
which, if successful, finalizes the inductive step.
You can write the left-hand side of (2) as
$$
\prod_{k=1}^{j}(I+u^{(k)}e_k^T)=\left[\prod_{k=1}^{j-1}(I+u^{(k)}e_k^T)\right](I+u^{(j)}e_j^T).
$$
But wait, we know what the term in the square brackets is equal to thanks to the assumption (1). So let's plug it in:
$$
\begin{split}
\prod_{k=1}^{j}(I+u^{(k)}e_k^T)&=\left[\prod_{k=1}^{j-1}(I+u^{(k)}e_k^T)\right](I+u^{(j)}e_j^T)
=
\left(I+\sum_{k=1}^{j-1}u^{(k)}e_k^T\right)(I+u^{(j)}e_j^T)\\
&=
I+\sum_{k=1}^{j}u^{(k)}e_k^T+\color{blue}{\sum_{k=1}^{j-1}u^{(k)}(e_k^Tu^{(j)})e_j^T}.
\end{split}
$$
Now given the assumed structure of $u^{(j)}$, can you show that the last expression is equal to the right-hand side of (2) (to make the blue term disappear)?
Remark: You can also use $n=1$ as the base case which makes its proof even easier.
