Why isn't $A=\lambda$ in $Au=\lambda u$? (eigenvalues) Why can't you divide both sides by the vector u? I don't understand how you can multiply the same vector by 2 different things and still get the same result.
 A: You're used to being able to divide by numbers. But for vectors that doesn't work. You can't divide by a vector.
If you try to define such an operation yourself, you'll quickly run into trouble. What should the result of dividing one vector by another be? Another vector? A real number? You're expecting $\lambda u / u$ to give you $\lambda$, which is a real number, but $Au / u$ to give you $A$, which is a matrix. Such an operation doesn't make any sense at all.
A: It can be hard to answer questions of the form "why can't you...". But regardless of the "why", a simple example shows that you simply can't.  Let $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$, $u=\begin{pmatrix}1\\1\end{pmatrix}$ and $\lambda=2$. Then $Au=\lambda u$ but $A\ne\lambda$.
A: Dividing both sides by the same thing doesn't always make sense in that the resulting answer is meaningless, e.g. dividing by $u$
$$
Au = \lambda u \Rightarrow A = \lambda
$$
i.e. a matrix = a scalar, which is obviously wrong.
Other examples in other areas may be constructed
$$
\frac{d}{dx}e^{kx} = k\,e^{kx} \Rightarrow \frac{d}{dx}=k
$$
i.e. a differential operator = a number.
Beware though, even if the correct answer is obtained, it does not imply the process of dividing/cancelling was valid ... cancel the $6$ in numerator and denominator
$$
\frac{16}{64} = \frac{1}{4}
$$
A: Because division by vectors is an undefined operation in general.
A: You cannot divide both sides of your expression by $u$ because dividing by vectors is not a well-defined operation, in general. 
In parallel, take for example the $2\times 2$ identity matrix $I$. If $u = (x,y)$, we shall have:
\begin{eqnarray}
Iu = \begin{pmatrix}
1 & 0 \\
0 & 1 
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
= \begin{pmatrix}
x \\
y
\end{pmatrix}
= 1 u
\tag{1}\label{1}
\end{eqnarray}
This is a trivial example, but it shows that it is possible to multiply a matrix by a vector and have, as a result, the vector multiplied by a scalar. The point in solving your equation $Au = \lambda u$, that is, the point on finding the eigenvalues and eigenvectors of a given matrix $A$ can be thought as: for what vector $u$ and constant $\lambda$ do we have $Au = \lambda u$ in the same way we had in (\ref{1})? 
In short: I believe you agree with the trivial case (\ref{1}), which shows that it is possible to have "two different things multiplying one thing and given the same result" as you said. On the other hand, there is no reason for us to believe this is the only case where something like this happens. The point, then, becomes finding other cases where this kind of property occurs and this is precisely what we are doing when we find the eigenvalues and eigenvectors for a given matrix $A$ of a given order and with respect to a given basis. 
