Why can't this matrix be diagonalized by its own eigenvector matrix?

Question

I have this matrix A diagonalized to the following eigen vectors:

But forming the following matrix B:

And then finding its inverse $B^{-1}$:

I'm unable to diagonalize A with $BAB^{-1}%$:

Why is this the case? This is just so baffling when it's supposed to work..

Answer 1

From $$AB=BD$$ we get $$B^{-1}AB =D$$

You have it the other way around.

Check the new $$B^{-1}AB =D$$ see if it works out.

Answer 2

The matrix $$ M=\begin{pmatrix} 2&-2\\ 1&4 \end{pmatrix} $$ is diagonalizable as $$ M=BDB^{-1} $$ where $D$ is the diagonal matrix of the eigenvalues $$ D=\begin{pmatrix} 3-i & 0\\ 0 & 3+i \end{pmatrix} $$ and $B$ is the matrix that has as columns the corresponding eigenvectors $$ B=\begin{pmatrix} -1-i & -1+i\\ 1 & 1 \end{pmatrix} $$

so we have also $$ B^{-1}MB=B^{-1}(BDB^{-1})B=D $$