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

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?

• Other than whether the diagonal matrix is called $D$ or $\Lambda$ there seems to be no difference between your first and last equations. – Henning Makholm Jan 23 '18 at 20:17

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 $$A$$.