Correct answer: (B) If S²=S and ranks agree, T and S are similar.; (D) T is necessarily diagonalizable.
Explanation
Minimal polynomial of T divides x(x−1), so T is diagonalizable with rank(T) copies of 1 and the remaining copies 0. This proves D and the rank criterion B for another projection. A fails since a nonidentity projection has a kernel. Eigenvalue sets alone do not fix multiplicities or Jordan blocks, refuting C. **B,D**.
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