Why does eigendecomposition is in the same form of $A=CBC^{-1}$ as using eigenvectors as basis for a transition matrix?

Question

Let A be a linear transformation matrix f(x)=Ax, and if D is A in the eigenvector basis, $$D = Q^{-1}AQ$$, where Q is eigenvector matrix of A and a change of basis matrix changing A in standard basis to a basis of eigenvectors of A. It seems to be similar to eigendecomposition $$ A = Q\Lambda Q^{-1} $$ $$ \Lambda = Q^{-1}AQ$$ So the linear transformation matrix D with eigenvectors of A as basis is just the eigenvalues matrix of A. What are the connections here?

Answer 1

You simply have different cases of the same result:

If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.

Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.

If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$

Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $\Lambda$ that is given by the same formula $\Lambda=Q^{-1}AQ$ (and $A=Q\Lambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.