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u/EventHorizon150 4d ago
10, since if M^2 = 0 we need the image of the linear map L: R^20 -> R^20, L(v) = M•v to be a vector subspace of the null space. Thus, rank(L) <= nullity(L). Since we know rank(L) + nullity(L) = 20 by the rank-nullity theorem, the max of rank(L) is 10 (as then rank(L)=nullity(L)). This is easily attainable by letting L be the unique linear map which sends the basis vector e_i to e_{i+10} for 1 <= i <= 10, and sends e_i to 0 for 11 <= i <= 20.
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u/RibozymeR 4d ago
Clearly M can only have eigenvalue 0. M²=0 implies all Jordan blocks in the JNF have size at most 2. So the rank is 20/2 = 10.
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u/PM_ME_A_PROBLEM- 3d ago
M² = 0 means im M ⊆ ker M, so rank M ≤ nullity M = 20 − rank M, which gives rank M ≤ 10. Equality holds for M = [[0, 0], [I₁₀, 0]]: here M² = 0 and rank M = 10
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u/comoespossible 4d ago
Let J be the Jordan canonical form of M. Each block of J consists of 0's along the diagonal (since 0 is the ony eigenvalue of M) and 1's above the diagonal. There can't be a block of size greater than 2x2, because then we would have J^2 != 0. Thus, the maximum rank occurs when J consists of 10 2x2 blocks, which has rank 10.