the index of the matrix (i.e., the smallest power after which null spaces stop growing). Examples. I've tried various things like assigning the matrix to variable A then do a solve(A^X = 0) but I only get "warning solutions may have been lost" 0 0 0 3 0. Theorem (Characterization of nilpotent matrices). For example, every [math]2 \times 2[/math] nilpotent matrix squares to zero. This definition can be applied in particular to square matrices.The matrix = is nilpotent because A 3 = 0. Nilpotent operator. 0 2 0 0 0. In general, sum and product of two nilpotent matrices are not necessarily nilpotent. I = I. Deï¬nition 2. The concept of a nilpotent matrix can be generalized to that of a nilpotent operator. But if the two nilpotent matrices commute, then their sum and product are nilpotent as well. 6 0 0 0 0. Recall that the Core-Nilpotent Decomposition of a singular matrix Aof index kproduces a block diagonal matrix â C 0 0 L ¸ similar to Ain which Cis non-singular, rank(C)=rank ¡ Ak ¢,and Lis nilpotent of index k.Isitpossible of A.The oï¬-diagonal entries of Tseem unpredictable and out of control. We highly recommend revising the lecture on the minimal polynomial while having the previous proposition in mind. A^m=0 may be true for just m=3 but not for m=1 or m=2. An n×n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. See nilpotent matrix for more.. But then 0 = CB^n = B^(n-1), a contradiction. Consequently, a nilpotent matrix cannot be invertible. The matrix A would still be called Nilpotent Matrix. nilpotent matrix The square matrix A is said to be nilpotent if A n = A â¢ A â¢ â¯ â¢ A â n times = ð for some positive integer n (here ð denotes the matrix where every entry is 0). A nilpotent matrix cannot have an inverse. As to your original problem, you know B^n = 0 for some n. The index of an [math]n \times n[/math] nilpotent matrix is always less than or equal to [math]n[/math]. In the factor ring Z/9Z, the equivalence class of 3 is nilpotent because 3 2 is congruent to 0 modulo 9.; Assume that two elements a, b in a ring R satisfy ab = 0.Then the element c = ba is nilpotent as c 2 = (ba) 2 = b(ab)a = 0. Say B^n = 0 where n is the smallest positive integer for which this is true. The determinant and trace of a nilpotent matrix are always zero. If, you still have problem in understanding then please feel free to write back. Products of Nilpotent Matrices Pei Yuan Wu* Department of Applied Mathematics National Chiao Tung University Hsinchu, Taiwan, Republic of China Submitted by Thomas J. Laffey ABSTRACT We show that any complex singular square matrix T is a product of two nilpotent matrices A and B with rank A = rank B = rank T except when T is a 2 X 2 nilpotent matrix of rank one. (b) Nilpotent Matrix: A nilpotent matrix is said to be nilpotent of index p, (p â N), i f A p = O, A p â 1 â O, \left( p\in N \right),\;\; if \;\;{{A}^{p}}=O,\,\,{{A}^{p-1}}\ne O, (p â N), i f A p = O, A p â 1 = O, i.e. Then CB = I. if p is the least positive integer for which A p = O, then A is said to be nilpotent of index p. (c) Periodic Matrix: This means that there is an index k such that Bk = O. By Nilpotent matrix, we mean any matrix A such that A^m = 0 where m can be any specific integer. Now suppose it were invertible and let C be it's inverse. It does not mean that A^m=0 for every integer. Hi, I have the following matrix and I have to find it's nilpotent index... 0 0 0 0 0. 0 0 8 0 0.

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