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?

In order to avoid confusion between scalar and vector multiplication, I use the notation $\langle x,y\rangle$ to denote the inner product $x\cdot y$ between vectors $x$ and $y$ in the following.
If $x$ is an eigenvector of $A$ to the eigenvalue $\lambda$, then $Ax = \lambda \cdot x$. Taking the inner product with $x$, this implies $\langle Ax,x\rangle= \lambda \cdot \langle x, x\rangle$, which is a scalar. Hence $$\frac{\langle Ax,x\rangle}{\langle x,x\rangle} = \frac{\langle \lambda\cdot x, x\rangle }{\langle x,x\rangle}= \frac{\lambda\cdot\langle x,x\rangle}{\langle x,x\rangle} = \lambda.$$ 
Note that in your argument you divide by the vector $I x$, which is something you cannot do! Taking the inner product with $x$ circumvents this issue, while preserving your intuition of the proof.
