Proof for corresponding eigenvalue If x is an eigenvector of a matrix A, then show that its corresponding eigenvalue is given by $\lambda=\dfrac{Ax\cdot x}{x\cdot x}$
I tried starting from $(A-\lambda{I})x=0$.
$Ax-\lambda{Ix}=0$
$\lambda{Ix}=Ax$
$\lambda=\dfrac{Ax}{Ix}$. This now is a bit confusing. Any help?

Your intuition starts off right, it is a good idea to begin with the definition of an eigenvalue, i.e. $A x = \lambda x$. Now the problem is that on both sides of the equation you have vectors, so you cannot divide by $x$. The idea is to "transform" the equation into an equation that only involves scalar quantities. A good way to do this is to take the scalar product with another vector $y$, i.e. (I will write the scalar product $x \cdot y$ as $\langle x, y \rangle$ to avoid confusing it with the normal multiplication of real numbers)
$$
\langle Ax, y \rangle = \lambda \langle x,y \rangle.
$$
Now the question arises: which $y$ should we choose? To solve the equation for $\lambda$, we obviously would like to divide by $\langle x,y \rangle$ (which is a real number now!). But this number might be zero if we choose the wrong vector $y$ (e.g. if $y = 0$ or $y \perp x$). So we need to make sure that $\langle x,y \rangle \neq 0$. The only thing we know is that $x$ is an eigenvector. But by definition this means that $x \neq 0$, and by the properties of scalar products, you know that $\langle x, x \rangle \neq 0$ if and only if $x \neq 0$. 
So $y = x$ looks like a good choice!
Then we get
$$
\langle Ax, x \rangle = \lambda \langle x,x \rangle \Rightarrow \lambda = \frac{\langle Ax, x \rangle}{\langle x,x \rangle}
$$
or in your notation
$$
\lambda = \frac{Ax \cdot x}{x \cdot x}.
$$
