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 $b_2 = \begin{bmatrix}-1\\3 \end{bmatrix}$.

  • $\begingroup$ The bases were given as part of the problem. $\endgroup$ – Kot Aug 15 '14 at 1:56
  • $\begingroup$ If you use two different bases, you're going to get two different matrices. The first basis in your post does not diagonalize $A$, and the second basis vector in that basis is not an eigenvector. $\endgroup$ – David Wheeler Aug 19 '14 at 10:43

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