Find basis so that positive definite matrix acts like identity

Question

I am trying to see if it is possible to find a basis such that a positive definite matrix $A$ acts like the identity matrix on that basis. I know from the spectral theorem $A=X\Lambda X^{-1}$ for eigenvectors $X$ and eigenvalues $\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n)$. From here, it is clear that if I start with the relation $y = Ax$ then this implies $X^{-1}y = \Lambda X^{-1} x$ so taking $\tilde{y} = X^{-1} y$ and $\tilde{x}=X^{-1} x$ gives a basis expansion such that $\tilde{y} = \Lambda \tilde{x}$. So it is at least possible to find a basis for which $A$ acts like a diagonal matrix. I'm having trouble showing (if it is even possible) to extend this to find a basis for which $A$ acts like the identity matrix. Can I have some help?

Answer 1

If there exists a basis where a matrix acts as the identity (meaning it leaves the elements of the basis unchanged) then the matrix is the identity (because every vector is a combination of elements of the basis). So what you're trying to do never works unless the matrix is actually the identity.

Answer 2

This post seems to be confusing two different notions of 'finding a basis' or a 'change of basis' and says things like "such that a positive definite matrix acts like the identity the identity matrix on that basis", which seems rather vague.

If we are interested in $A$ as a linear operator, then this is impossible unless it is already the identity matrix.

If we are interested in bilinear forms, which are at the heart of the definition of a positive definite matrix, then our hermitian positive definite $A$ certainly is congruent to the identity matrix, which is what Sylvester's Law of Inertia tells us. E.g. using Cholesky factorization we have $A=LL^*$ so $\big(L^{-1}\big)A\big(L^{-1}\big)^*=I$ is a change of basis with respect to a bilinear form that gets $A$ 'to act like the identity matrix'