How to explain that I is needed for characteristic equation? 
$$\\A\mathbf v - \lambda \mathbf v=0$$
  $$\\(A - \lambda) \mathbf v=0$$

since the equation looks like to be correct. how to explain that it is not? (without examples) 
I know that$\ I$  makes no change for $\lambda \mathbf v$ and
i only found unpersuasive explains like this one "We needed to do this because without it we would have had the difference of a matrix, A, and a constant, λ, and this can’t be done." can't we subtract a scalar from a matrix?
 A: What you put in quotes is the explanation. Why try to do this without an example? $2 \times 2$ should be convincing, particularly if you note the ambiguity: perhaps "subtracting  $\lambda$" should mean subtracting it from all four matrix entries, not just from the two on the diagonal. How would you distinguish?
A: The size of $A$ and $\lambda$ are not the same, hence the subtraction is not well defined. 
While you can define $A- \lambda = A - \lambda I$, until you do so, the subtraction is not compatible and it might not be clear to other readers of what you are doing. We do call a diagonal matrix with identical diagonal entries scalar matrix but I am not aware of the convention to drop the $I$ term. A problem which could arise is I might have difficulty answering what is the size of $\lambda$. 
A: I guess you're thinking of "substracting a scalar from a matrix" as "substracting a scalar from each element of a matrix". Let's call $\star$ this operation. Notice that $A-\lambda I \neq A\star \lambda$.  For instance, with $A = I$ and $\lambda = 1$ : 
$$ A - \lambda I = I - 1\times I = 0$$
while
$$A \star \lambda = I\star 1 = \pmatrix{1-1 & 0 -1 \\ 0 -1 & 1-1} = \pmatrix{0 & -1 \\ -1 & 0}$$
Thus the equation $Av - \lambda v = 0$ is not equivalent to $(A\star \lambda) v = 0$ ; but it is equivalent to $(A-\lambda I)v = 0$.
A: The field of scalars embeds into the matrix ring via $\lambda \mapsto \lambda I$, therefore you can identify $\lambda \in \Bbb{k}$ the field with $\lambda I$.  Thus the notation is well-defined.
