Can every diagonalizable matrix be diagonalized into the identity matrix? I'm a chemistry major and I haven't taken much math, but this came up in a discussion of quantum chemistry and my professor said (not very confidently) that if a matrix is diagonalizable, then you should be able to diagonalize it to the identity matrix. I suspect this is true for symmetrical matrices, but not all matrices. Is that correct?
 A: No.  If $PAP^{-1} = I$ where $I$ is the identity then $A = P^{-1}IP = P^{-1}P = I$.  So in fact only the identity matrix can be diagonalized to the identity matrix.
A: Take the $0$ $n\times n$ matrix. It's already diagonal (and symmetrical) but certainly can't be diagonalized to the identity matrix. 
A: The usual meaning of "diagonalization" is diagonalization by similarity transform, which takes the form of $PAP^{-1}=D$. As the others have shown, the only matrix that can be diagonalized into the identity matrix is the identity matrix itself.
Yet, depending on the context, your professor may refer to diagonalization by congruence, which takes the form of $P^\ast AP=D$. If he implicitly assumes that $A$ is positive definite, then his assertion is true: $P^\ast AP=D\,\Rightarrow\,(PD^{-1/2})^\ast A(PD^{-1/2})=I$. However, since I know absolutely nothing about quantum chemistry, I can't say if he really meant this.
A: I know that the question has already been answered, but I hope that the following line of thought will provide further insight nonetheless:
The trace of an n-dimensional Hermitian (because we want a guarantee that it has n eigenvalues) matrix is a basis invariant given by
a) the sum of the eigenvalues
b) the sum of the diagonal elements.
Consequently, if for every Hermitian matrix, there existed an orthogonal transformation capable of diagonalizing it to the identity matrix, we would equivalently have that
a) the sum of the the eigenvalues of every n-dimensional Hermitian matrix is n
b) the trace of every n-dimensional Hermitian matrix is n,
both of which are trivially disproven by counter-example. Even more disturbingly, since we are searching for a transformation that diagonalizes the matrix, we know that the we would have, in addition, that
c) every n-dimensional Hermitian matrix has only the n-fold degenerate eigenvalue 1.
and computational quantum chemistry would be very dull indeed.
