Help with understanding similar matrices Going over past notes I do not understand the concept of similar matrices and fail to see how my lecturer has got these Eigenvectors associated to the values, I think I'm missing something can someone help in layman terms.

 A: I will give you some pointers, however I strongly suggest you read about the subject in a source that is independent from your lecture, because getting a second perspective can be better than brooding over course material.
If you have a triangular matrix, the eigenvalues can be read from the diagonal; this follows from considering the characteristic polynomials, I will not go into detail here. In general consider the equation $Av = \lambda v$ where $A$ is a matrix, $v\neq 0$ is a vector and $\lambda\neq 0$. If such an equation holds, then $v$ is called an eigenvector and $\lambda$ is called an eigenvalue of $A$ (this is the definition of these terms). Note that $v\neq 0$ is key for this to make sense, but you can have $\lambda =0$. So If you know an eigenvalue $\lambda$, you can compute an associated eigenvector by solving $Av = \lambda v$ which is equivalent to $(A-\lambda I)v = 0$. This is what your instructor does to obtain the eigenvectors. 
The aim now is to diagonalize the matrix and to do this one uses the eigenvectors. I will explain why $P^{-1}AP$ is diagonal, since you might not have seen why. Put the eigenvectors $v_1, \dots, v_n$ (that correspond to the eigenvalues $\lambda_1, \dots, \lambda_n$) as column vectors in a matrix denoted by $P$ (perhaps $n=3$ in your case). Also let $e_1, \dots, e_n$ be the standard basis. Since the k-th column of $P$ is $v_k$, we have $Pe_k = v_k$ and $P^{-1}v_k = e_k$ for any k. Remember $Av_k = \lambda_k v_k$ since these are eigenvalue-eigenvector pairs. So 
$$P^{-1}AP e_k = P^{-1}A v_k = P^{-1}\lambda_k v_k = \lambda_k e_k.$$
This shows that the k-th column of $P^{-1}AP$ is $\lambda_k e_k$, so it is diagonal with $\lambda_k$ in the k-th entry.
From the calculation you can see that reordering the $v_k$ will change the order in which the $\lambda_k$ appear on the diagonal, but not much else. Obviously, the values $\lambda_k$ on the diagonal are the eigenvalues of the original matrix. There's still more to be said here like what happens when you have multiple linear independent eigenvectors for the same eigenvalue and why $P^{-1}$ is invertible, but in your example these are no concern.
