Diagonal Matrix, just eigenvalues? Assuming I've tested for diagonalization, can I just take the eigenvalues and arbitrarily place them in in the i,j cells to produce a diagonal matrix?
Say I have a matrix $M$ with eigenvalues $\lambda_1 = 4,\; \lambda_2 = \lambda_3 = -2.$
$$M =\begin{bmatrix}{1} & {-3} & {3} \\ {3} & {-5} & {3} \\ {6} & {-6} & {4}\end{bmatrix}$$
And let,
$$A = \begin{bmatrix}4&0&0\\0&-2&0\\0&0&-2\end{bmatrix} , \quad B = \begin{bmatrix}-2&0&0\\0&4&0\\0&0&-2\end{bmatrix} , \quad C = \begin{bmatrix}-2&0&0\\0&-2&0\\0&0&4\end{bmatrix}$$
Are $A , B , C$ valid diagonal matrices , or does the order of the eigenvalues matter?
 A: Yes.  Assuming that your matrix is in fact diagonalizable (which will happen if all of the eigenvalues are distinct, but can also sometimes happen when you have repeated eigenvalues), then your matrix will be similar to ANY diagonal matrix that has the eigenvalues (with proper multiplicities) along the diagonal.
One way to see this is to look at what happens when you conjugate a matrix by a permutation matrix, that is, a matrix where every row and every column has exactly one nonzero entry, and that entry is equal to 1.  Doing so just swaps rows and columns around (and doesn't change the values of the entries of the matrix), and does so in a way that are along the diagonal remain on the diagonal.
If you play around with conjugation by permutation matrices, you should be able to come up with an explicit way to conjugate $\pmatrix{a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c}$ into $\pmatrix{c & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b}$, for example, and if you can figure out that, you will see how every diagonal matrix with a given set of entries is similar to each other.
A: Yes.
Let $A : V \to V$ be a linear transformation that has diagonal matrix representation
 $$D = 
\left(
 \begin{array}{ccccc}
   \lambda_1 & & \huge0 \\
    & \ddots  & \\
   \huge0 & & \lambda_n \\
 \end{array}
\right)$$ 
with respect to some basis $e_1, ..., e_n$. Then for all $k$, $$ D e_k  = \lambda_k e_k$$ that is each $\lambda_k$ is an eigenvalue.
Furthermore the ordering of the $\lambda$s is determined by the ordering of the basis which is arbitrary.
