Question
Question: A square matrix A is said to be nilpotent of index m. If \[{A^m} = 0\], now, if for this A,\[{\left(...
A square matrix A is said to be nilpotent of index m. If Am=0, now, if for this A,(I−A)n=I+A+A2+A3 +....+Am−1, then n is equal to?
Solution
Here we are using the binomial theorem to expand expressions of the form (I−A)n. After expanding we compare this equation with the given equation to find the value of n. We compare the coefficient of each term of both the equation to find the value of n.
Complete step-by-step answer:
It is given that A is a square matrix which is nilpotent of index m ifAm=0
For this A, (I−A)n=I+A+A2+A3 +....+Am−1
Am=0 .... (1)
So Am+1=0, Am+2=0, ……Am+n=0
It turns implies that
(I−A)n=I+A+A2+A3 +.....+Am−1…… (2)
Now, we use the binomial theorem for expansion
(I−A)n=nC0I+nC1(−A)1+nC2(−A)2+nC3(−A)3+……
(I−A)n=I−nA1+nC2A2−nC3A3+…… (3)
Now, we compare equation (2) and (3) to find the value of n and we get
− n = 1
n = − 1
Therefore, if n=−1 then (I−A)n=I+A+A2+A3 +....+Am−1.
Note: The Binomial theorem tells us how to expand expressions of the form. The binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. In linear algebra, a nilpotent matrix is a square matrix N such that Nk=0. For some positive integer k. The smallest integer such k is called the index of N, sometimes the degree of N.
A square matrix is a matrix with the same number of rows and columns. An n-by-n matrix is known as a square matrix of order.
An example A = \left[ {\begin{array}{*{20}{c}}
0&1 \\\
0&0
\end{array}} \right] is nilpotent with index 2, since A2=0.