Eigenspaces and Diagonal Matrices

Diagonal matrices. Eigenspaces. Conditions equivalent to diagonalizability.

[hong-yeopsong]

Here, an eigenspace of lambda and T is defined for any number lamba, whether it is an eigenvalue or not. Therefore, when lambda is NOT an eigenvalue, the correponding eigenspace is simply {0} and dos not contain any eigenvector. When lambda is an eigenvalue of T, then the eigenspace is not trivial and contains some non-zero vectors all of which are in fact eigenvectors of T correponding to lambda.

[twisthz]

Dear Prof. Axler - in this video and your book pg.156-157, I understand the key assumption we are making is *distinct* eigenvalues. However, I read elsewhere that linear operators can be diagonalized even if there are nondistinct/repeated eigenvalues, provided that, for each repeated eigenvalue (algebraic multiplicity), there are as many eigenvectors within the basis of the corresponding eigenspace (geometric multiplicity), so that consequently V still has a basis of eigenvectors (that is, there are enough eigenvectors to span V).

What I would like to check is that, in the case there are nondistinct/repeated eigenvalues, the sum of eigenspaces is still a direct sum. This is because the eigenvectors constituting the basis of any given eigenspace (of a repeated eigenvalue) are linear independent, in addition to the eigenvectors of distinct eigenvalues being linearly independent. So even though we would have fewer (that is, less than m) eigenspaces in total, the vector space V would still be a direct sum of those eigenspaces, and the dimensions of those eigenspaces added up would equal the dimension of V. Thank you very much.

[marcourielmedinamandujano]

why does the linear independency of $u_1, ..., u_n$ and $u_1+...+u_n=0$ imply $u_j=0$?