The Diagonal Matrix Representation Theorem states:
Suppose $A=PDP^{-1}$, where $D$ is a diagonal $nxn$ matrix. If $B$ is the basis for $R^n$ formed from the columns of $P$, then $D$ is the $B$-matrix for the transformation $x\to Ax$.
I am given matrix $A= \begin{bmatrix} -4 & -1\\ 6 & 1 \end{bmatrix} $ and $b_{1}= \begin{bmatrix} -1\\ 2 \end{bmatrix} $, $b_{2}=b_{1}= \begin{bmatrix} -1\\ 1 \end{bmatrix}$
In an example from class the $B$-matrix is found like this:
$Ab_{1}=\begin{bmatrix} 2\\ -4 \end{bmatrix}$ Solving for the $B$-coordinate vector gives $[T(b_{1})]_{B} =\begin{bmatrix} -2\\ 0 \end{bmatrix}$
$Ab_{2}=\begin{bmatrix} 3\\ -5 \end{bmatrix}$ Solving for the $B$-coordinate vector gives $[T(b_{2})]_{B} =\begin{bmatrix} -2\\ -1 \end{bmatrix}$
Combining these two vectors give the $B$-matrix $ \begin{bmatrix} -2 & -2\\ 0 & -1 \end{bmatrix} $
According to the Diagonal Matrix Representation Theorem, I should be able to get the same matrix if I diagonalize matrix $A$. $A$ has two eigenvalues $\lambda=-2,-1$. I have found the basis for each eigenspace.
For $\lambda = -2$, the basis is $\begin{bmatrix} -1\\ 2 \end{bmatrix}$ For $\lambda = -1$, the basis is $\begin{bmatrix} -1\\ 3 \end{bmatrix}$
So $P=\begin{bmatrix} -1 & -1\\ 2 & 3 \end{bmatrix}$ and $D= \begin{bmatrix} -2 & 0\\ 0 & -1 \end{bmatrix}$
The theorem states that $D$ is the $B$-matrix of the transformation but these two matrices are not the same. When I diagonlized $A$, I found the basis for each eigenspace and formed matrix $P$ with the vectors in the basis, and the corresponding eigenvalue of each vector forms matrix $D$. The only difference between the two matrices is the $-2$ in the top right corner. When diagonalizing a matrix shouldn't $D$ only have nonzero entries on it's main diagonal by definition? How would I get a $-2$ in the top right corner of $D$. Am I misinterpreting the theorem?
You are using two different bases. Try using $b_2 = \begin{bmatrix}-1\\3 \end{bmatrix}$.