#CSIRNET MATHEMATICS NOV 2020 | 30th Nov | Section C | Linear Algebra | Nilpotent Matrix

Hello friends, this is Mahesh Yovlik and you are watching Next Gen Math. If you have not subscribed my channel, then please subscribe the channel and press the bell icon for the updates of new videos. In this video, we are going to discuss the problem on CSI 8, November 2020, Problem of Linear Algebra. Question, let A be a 3 cross 3, nilpotent matrix, which of the following statements are necessarily true? We have given four options. First option, i plus a power n equal to i, for sum n greater than 0, where i is the identity matrix. Second option, the column space of A is singleton 0. Third option, the eigenvalues of A are roots of 1. And fourth option, A cube is diagonalizable matrix. If you want notes of this solution, then please visit our website www.nextgenmath.com. In the description, link is available. It is a problem of nilpotent matrix. So, first we recall the definition of nilpotent matrix. Matrix is said to be nilpotent matrix if there exists positive integer n such that a power n equal to 0 and such smallest n is called index of A. So, for the nilpotent matrix, there should be exist an integer n such that a power n equal to 0 and such smallest n is called the index of A. So, we will check all the option one by one. So, option one, we use the counter example. Example is A equal to 0 0 1 0 0 0 0 0 0. A square equal to 0 0 0 0 0 0 0 0 0. That means A square equal to 0. So, A is nilpotent matrix with index 2 because 2 is the smallest number such that A square equal to 0. Now, I plus A. So, I is identity matrix 1 0 0 0 1 0 and 0 0 1. Add it to A, then we have 1 0 1 0 1 0 0 0. Now, if we take n power of I plus A will be 1 0 and 0 1 0 0 0 1. That means if n equal to 2, then matrix will be 1 0 2 0 1 0 0 0 1. If n equal to 3, then matrix will be 1 0 3 0 1 0 0 0 1. But, this never be A identity matrix. But, the first option says that I plus A power n equal to I for some n greater than 0 where I is the identity matrix which is wrong. Now, option number 2. We take the same example. Now, the column space of A is the space generated by the column vectors and here the column vectors are 0 0 0 0 0 0 and 1 0 0. Therefore, the column space of this matrix is alpha times 1 0 0 where alpha belongs to R and which is not a singleton 0. So, option number 2 is also the wrong answer. Now, option number 3. For that, we have one property of nilpotent matrix. All the eigenvalues of nilpotent matrix are 0. But, in the option number 3, we have given that they should be the roots of 1. But, roots of 1 cannot be 0. So, option number 3 is also wrong answer. Now, for option number 4, we have another property of nilpotent matrix. The index of n n cross n nilpotent matrix is always less than or equal to n. That means, here the matrix is of order 3 cross 3. Therefore, the index of n 3 cross 3 nilpotent matrix is always less than or equal to 3. Hence, A cube equal to 0 because n equal to 3 and the index is less than 3. So, A power 3 must be 0 which is diagonalizable matrix. 0 matrix is always diagonalizable matrix. So, option number 4 which says that A cube is diagonalizable which is correct option. Thanks for watching. For more videos, please like, share, subscribe. Press the bell icon of NextGenMaths.