Transformation Matrix Problem So  I am a bit foggy with transformation matrices, and have an attempted solution.
 I am not sure how to double check my result however.
Let S be the ordered basis S = $\{e_1, e_2, e_3\}$ and let $|$T$_s$$|$ = \begin{bmatrix} 1 & 3 & 0 \\ 3 & 1 & 4 \\ 0 & 4 & 1 \end{bmatrix} 
; and let B $\{ (-4,0,3)^{T}, (3,-5,4)^{T} , (3,5,4)^{T} \}$ be the second basis. The question asks find [ T($\alpha_i$)]$_B$.
Now my equation I am using is $[T\alpha]_B$ = [ $I_{BS}]$$[T\alpha]_S$  
So my $[I_{SB}]$ = \begin{bmatrix} -4 & 3 & 3 \\ 0 & -5 & 5 \\ 3 & 4 & 4 \end{bmatrix} 
$[I_{BS}]$ = \begin{bmatrix} -4/25 & 0 & 3/25 \\ 0 & 12/50 & -16/50 \\ 18/50 & 30/50 & 24/50 \end{bmatrix}
so for example my [$T\alpha_1]$$_S$ = $\begin{bmatrix} -4 \\ 0 \\ 3 \end{bmatrix} $
SO I just multiplied $[I_{BS}]$$\cdot$[$T\alpha_1]$  to give the first column of [$T\alpha_B]$.  and thus repeat.
But the second part of the question asks for one to find $P^{-1}|T_s|P $ = diag(-4,6,1). 
Now I'm not sure how to do this.  but I have found the following :
the multiplication of $|$T$|$$_S$ $\cdot$ $B_1$ ( the first column vector $\begin{bmatrix} -4 \\ 0 \\ 3 \end{bmatrix}$ yields the coordinate vector (1,0,0).  
If I do the same for each vector in B, then the second coordinate vector is (0,-4,0), and the third (0,0,6).  
which is the same diagonal as the result I need.  Can someone help me understand what this means, and how to solve the last part of the problem finding P; and confirm if my first result is the correct way of thinking ?
 A: It is difficult to comment on the first part of your question, because it is not clear where the vectors $\alpha_i$ come from. Perhaps $\alpha_i=e_i \in S$? Or perhaps the $\alpha_i$ are the vectors in $B$? Also, I am not that familiar with the notation you are using, but here goes...
Note that the matrix representation of the transformation $T$ is always relative to two bases (which could be the same) - one basis is for the domain and one for the codomain of $T$ (again these could be the same, but they might have different bases). So I'm presuming $|T_S|$ is Mtx$_{S,S}(T)$ and column $i$ of this matrix is $[T(e_i)]_S$. So presuming your $\alpha_i$ is the $i$th vector in $B$, then $[T(\alpha_i)]_B$ would be the i'th column of Mtx$_{B,B}(T)$ and to get to that from Mtx$_{S,S}(T)$ you would need to do the following: \begin{equation} \text{Mtx}_{B,B}(T)=\text{Mtx}_{B,S}(\iota)\text{ Mtx}_{S,S}(T) \text{ Mtx}_{S,B}(\iota). \end{equation}So Mtx$_{B,S}(\iota)$ is your $I_{BS}$, and so on...so then, no, unfortunately it seems as if the first part of your question is not correctly answered the way you approached it. If you were just changing a vector from one basis representation to the other, that would work, but the question is asking you to change the matrix representation of the transformation relative to one basis (domain and codomain) to another (domain and codomain). So the main error I see is that you have $[T(\alpha_1)]_S=\begin{bmatrix} -4\\0\\3 \end{bmatrix}$. This is incorrect, it is actually $[\alpha_1]_S=\begin{bmatrix} -4\\0\\3 \end{bmatrix}$. I hope that helps.
For the second part - this is a diagonalization problem - so the matrix $P$ has as its columns the eigenvectors associated with each of the eigenvalues -4, 6 and 1. So you can read more about it on wikipedia (http://en.wikipedia.org/wiki/Matrix_diagonalization)...but essentially, since you are given the eigenvalues, you just need to solve the homogenous systems $\lambda_iI-|T_S|=0$ associated with each eigenvalue to find the eigenvectors.
