Implementation of Block LMS In the implementation of block LMS, i need one clarification.

In the 3rd step as shown in the figure attached, the summation over a product of input, $\mathbf{u}$ and error, $e$, associated with each sample value in a block, gives rise to a single value (some integer). 


Every first statement (formulae) in a line belongs to Block LMS algorithm and second statement (formulae) belongs to simple LMS Algorithm. What I want is only to consider first formulas in every line.

If all the initial weights $\hat{w}(k)$ are the same (all are zeros), then this constant value will be added to that at the end of each block which results in updates weights having the same value. i.e; 


  [0 0 0 0 0] + x = [x x x x x].


If this process continues till the end of input, all the filter weights will be of the same value. 

Is this correct?
As applesoup says in the comments the term
$$
\mathbf{u}(kL+i)e(kL+i)
$$
is a vector, not a single value (some integer).

Why do you think it's a scalar?

To answer your question: no, it's incorrect to state that all the algorithm does is add a constant to all elements of $\hat{w}(k)$.

As per this slide from here



the block LMS error is still a scalar. It's just that the errors across multiple blocks are used to form the updates.

Another way to look at it is that $\Phi(k)$ can be written as:
$$
\mathbf{u}_k \mathbf{e}_k
$$
where $$\mathbf{u}_k = 
\left[
\begin{eqnarray*}
&amp;\mathbf{u}(kL+0)
&amp;\mathbf{u}(kL+1)
&amp;\mathbf{u}(kL+2)
&amp;\ldots
&amp;\mathbf{u}(kL+L-1)
\end{eqnarray*}
\right]$$
is an $N_u \times L$ array and
$$\mathbf{e}_k = 
\left[
\begin{eqnarray*}
&amp;e(kL+0)\\
&amp;e(kL+1)\\
&amp;e(kL+2)\\
&amp;\vdots\\
&amp;e(kL+L-1)\\
\end{eqnarray*}
\right]$$
is an $L\times 1 $ vector.

As a result $\mathbf{u}_k \mathbf{e}_k$ is an $N_u \times 1$ vector.