Spectral theorem for diagonal matrix in different inner product spaces

Question

I learned a special case of the spectral theorem for finite dimensional inner product space. As I understand it states that a real matrix is orthogonally diagonalizable with real eigenvalues iff it equal its hermitian adjoint. However when I try to apply it to diagonal matrices, I think I have a problem. If I work with the regular inner product ($\langle u,v\rangle \to u^tv)$ then I see how this works out ($A^* = A^t = A$), but if I work in a more general inner product space: $\langle u,v\rangle = (Bu)^t(Bv)$ where $B$ is an invertiable matrix, then the hermitian adjoint of a matrix $A$ becomes $A^* = (B^tB)^{-1}A^t(B^tB)$, and this may not commute with $A$, if $A$ is some general diagonal matrix, for example let's define $A = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}, B = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$. However we still have that for the identity matrix: $I^* = I$ so a diagonal matrix, is still orthogonally diagonalizable. How this doesn't contradict the spectral theorem?

Answer 1

Sorry for the confusion. Here is the real answer.

The true statement of the spectral theorem is that

$A$ is selfadjoint iff there exists an orthonormal basis where $A$ is real diagonal.

notice that having an orthonormal basis is way different than saying $A=QDQ^*$ where $D$ is real diagonal and $Q$ is orthogonal.

Instead, it means that $A=MDM^{-1}$ where the columns of $M$ are orthonormal, i.e. $<m_i,m_j>=\delta_{i,j}$.

In fact, the columns of the identity matrix $I$ do not form a orthonormal basis since $IB^TBI \ne I$ , or also said $<e_i,e_j> = (B^TB)_{i,j}\ne \delta_{i,j}$.

Hope that this makes it clear for you.