Am I correct in assuming that 0 entry is the zero matrix?
I think is correct, bopbop. In the following I will take to be the original matrix, i.e. is the given block matrix with in the upper right corner. For generality, I'm assuming is an matrix. Here's an outline of what I think supports this:
1) An matrix is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to . See Diagonalizable matrix - Wikipedia, the free encyclopedia for this statement.
2) Since the original matrix is upper triangular with 1's along the main diagonal, the only eigenvalue is 1.
3) From 2) we know the only eigenspace is that corresponding to the eigenvalue 1, because there were no other eigenvalues. By definition, the eigenspace belonging to 1 is , where is the identity matrix. For to be diagonalizable, 1) tells us that we want the dimension of (i.e. the nullity of ) to be . The matrix is the block matrix with in the upper right corner and 's for the other entries. Now, if , then the rank of is at least one, because there will be at least one linearly independent column of when . By the rank-nullity theorem, this implies the nullity of is less than or equal to . Since we require the nullity of to be , we must have .
If something is unclear or incorrect let me know. Hope this helps, MisterF!