# Eigendecomposition of a matrix implies invertibility? [duplicate]

Does existence of Eigendecomposition of a matrix imply that it is invertible? That is, does the existence of n independent eigenvectors of a matrix imply that the matrix is invertible? If so, what is the reasoning behind eigenvectors and invertibility being connected? It seems intuitively wrong based on what I've learnt.

For instance, suppose A has n independent eigenvectors. Then the eigendecomposition exists:

$$A = PDP^{-1}$$

Then A is invertible?

EDIT: link to another question on this topic: For a square matrix $A$ presence of $n$ linearly independent eigenvectors implies full rank, thereby invertible?

• No. Any diagonal matrix has a full set of eigenvectors, but any of the diagonal elements can be zero leaving the matrix singular. – copper.hat Mar 26 at 3:07
• In particular, the $0$ matrix is diagonal, and so if you pick your favourite invertible matrix $P$, then $0 = P0P^{-1}$. – user759562 Mar 26 at 3:11
• @information_interchange Having an eigenvalue equal $0$ means that the matrix is not invertible. If there had to be a case, it has to correspond to the eigenvalue $0$. – EuxhenH Mar 26 at 3:18
• You can cook up cases of matrices having $n$ independent eigenvectors and rank $k$, for every $k=1,\dots,n$. – EuxhenH Mar 26 at 3:28
• Why does having a nontrivial kernel make diagonalization useless? – amd Mar 26 at 4:47