Diagonalizable iff $V$ is the direct sum of eigenspace

Question

Here comes the fundamental definition of diagonalizable : $A=PDP^{-1}$ And we can prove that $A$ is diagonalizable iff $A$ has $n$ linear independent eigenvectors. But here is another theorem saying that A is diagonalizable iff $V$ decomposes as a direct sum of eigenspaces, i.e. $V=V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_k}$ for $dim V=n$, but I am not sure whether the k here equals to n, because the basis of each eigenspace $V_{\lambda_i}$ may contain more than one element, so it seems that $k\leq n$. Or is it that for the diagonalizable cases, we have $n$ distinct eigenvalues, so the cardinality of basis of each eigenspace is exactly one? Could anyone help me? Thank you

Answer 1

If n distinct eigenvalues are there then diagonalizability follows. But matrices with repeated eigenvalues can also be diaganozibale.

If $V$ decomposes into direct sum of eigenspaces, then take any basis for each eigen space and then their union. This union will have exactly n elements and will consist of eigen vectors, proving diagonalizability. We can see that dimension of individual eigenspaces do not matter. The main thing all these eigen spaces together GENERATE the whole of V; in general they may not.