Treatment/Properties of the identity matrix when calculating eigenvectors When calculating eigenvectors, we algebraically manipulate the relationship between eigenvectors and eigenvalues:
$$ A\mathbf{v} = \lambda\mathbf{v} $$
$$ \Rightarrow A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$$
$$ \Rightarrow \mathbf{v}(AI -\lambda I) = \mathbf{0}$$
My confusion lies with how the identity matrix, $I$, is treated.
We know that $ \mathbf{v} $ is a $2$x$1$ matrix (a vector). In this case, the multiplicative identity of $ \mathbf{v} $ is the $1$x$1$ identity matrix, $I$. This is because we know that if we were to distribute $ \mathbf{v} $ in $ \mathbf{v}(AI -\lambda I) = \mathbf{0} $, we would need to get $ A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0} $ again. Therefore, the identity matrix must have rank $1$.
However, if this were true, in that the identity matrix is a $1$x$1$ matrix consisting of only the value $1$, then the expression $ AI -\lambda I $ would be invalid. This is because, $A$ is a $2$x$2$ matrix, $I$ is a $1$x$1$ identity matrix, and $\lambda$ is a scalar.
Given this reasoning, I suspect there must be something incorrect with my understanding of the identity matrix, its properties, and how these properties relate to this specific situation of eigenvectors and eigenvalues. I would greatly appreciate it if someone could please clarify my incorrect reasoning. Please state why my reasoning is incorrect, what the correct reasoning is, and an example to help illustrate your point.
Thank you.
EDIT
The users Eff and Jack were able to help me understand the errors in my reasoning. 
As Jack said in his answer, my error was in assuming commutativity of matrix multiplication, and then applying this to distribute the matrix through an expression.
This would be possible if $I$ was a $2$x$2$ matrix, since $I\mathbf{v} = \mathbf{v}$. However, $\mathbf{v}I \not= \mathbf{v}$. My confusion must lie in my not differentiating between the left multiplication (in this case, distribution) of a matrix ($A\mathbf{v} - \lambda\mathbf{v} = 0 \not\Rightarrow \mathbf{v}(AI - \lambda I) = 0 $) and the right multiplication (in this case, distribution) of a matrix ($A\mathbf{v} - \lambda\mathbf{v} = 0 \Rightarrow (AI - \lambda I)\mathbf{v} = 0$). 
$\therefore \mathbf{v}(AI - \lambda I) \not= (AI - \lambda I)\mathbf{v}$.
 A: In your reasoning

$$ A\mathbf{v} = \lambda\mathbf{v} $$
  $$ \Rightarrow A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$$
  $$ \Rightarrow \mathbf{v}(AI -\lambda I) = \mathbf{0}$$

the second line is correct but the third line is wrong simply because the sizes of the matrices do not satisfy the requirement for matrix multiplication. Even when sizes match, matrix multiplication is not commutative in general. 

We know that $ \mathbf{v} $ is a $2\times 1$ matrix (a vector). In this case, the multiplicative identity of $ \mathbf{v} $ is the $1$x$1$ identity matrix, $I$. 

This does not make sense. There seems to be confusion about the concept "multiplicative identity". For simplicity, I'm assuming $v\in\mathbb{R}^2$, which is a column vector. "the multiplicative identity of $v$" does not make sense. There is no such thing in linear algebra. A correct saying would be "the multiplicative identity of the ring of 2-by-2 real matrices". On the other hand, it true by definition that
$$
1v=v\tag{1}
$$
where $1$ is the scalar real number. But note carefully that one can not think $1$ as the 1-by-1 matrix $[1]$, because you cannot multiple a 1-by-1 matrix to a 2-by-1 matrix on the left since the size does not satisfy the requirement for matrix multiplication.

This is because we know that if we were to distribute $ \mathbf{v} $ in $ \mathbf{v}(AI -\lambda I) = \mathbf{0} $, we would need to get $ A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0} $ again. Therefore, the identity matrix must have rank $1$.

This argument does not make sense either. When you say "the identity matrix", you need to specify in mind. If it is the 2-by-2 matrix, then the rank must be $2$. If you are talking about $I$ being the 1-by-1 identity matrix, then the expression $AI-\lambda I$ would be completely wrong.
A: First off, it's a right-multiplication, i.e. 
$$Av - \lambda v = 0\implies (A-\lambda I)v = 0. $$
Secondly, the $I$ represents the idendity matrix of the same size as $A$, e.g. of size $2\times 2$ in your example. In your example
$$I = \left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right].
$$
In general, the idendity matrix $I$ is a diagonal matrix with ones in the diagonal and zeros everywhere else, and its size is usually obvious from the context.
A: If you do the matrix multiplication with $\mathbf{v}$ on the left, then, yes, in order to have $\mathbf{v} I=\mathbf{v}$, $I$ would have to be a $1\times1$ matrix. But, if you take $I$ to be $1\times1$, then you can't write $\mathbf{v}(AI -\lambda I)$ because $AI$ wouldn't make sense ($A$ being $2\times2$).
