CSIR NET MATHEMATICS| Linear Algebra | How to identify Diagonalizable Matrix? (Dec 2018)

In this video, we are going to discuss problem of linear algebra from CSIR NET, December 2018. Example number 66, booklet C, marks 4.75. A, B, real matrix with the characteristic polynomial x minus 1 power 3. Pick the correct statement from the below. First statement, A is necessarily diagonalizable. The second statement if the minimal polynomial of A is x minus 1 power 3 then A is diagonalizable. The characteristic polynomial of A square is x minus 1 power 3 and if A has exactly two Jordan blocks then A minus I square is diagonalizable. Now to answer this question we begin with the option one and to solve option we will use the counter example. So, first of all, matrix A is diagonalizable if its algebraic and geometric multiplicity are equal for each eigenvalue. Suppose A is 1 0 0, 1 1 0 and 1 1 1. Then the characteristic polynomial of A is determinant lambda I minus A is equal to determinant lambda minus 1 0 0 0 minus 1 lambda minus 1 0 and minus 1 minus 1 lambda minus 1. Its determinant value is lambda minus 1. So it satisfies the condition of the given problem A be a real matrix with the characteristic polynomial x minus 1 power 3. Moreover the eigenvalues are the roots of the characteristic equations. Therefore the eigenvalues are 1 1 1 for the matrix A. So algebraic multiplicity for lambda equal to 1 is 3. Now we are going to find the geometric multiplicity. That means we have to calculate the eigenvectors. So, eigenvectors for lambda equal to 1, A minus lambda I into x is equal to 0. So, lambda equal to 1, so we have A minus I into x is equal to 0. The solution of this homogeneous system of linear equation will give the eigenvectors for lambda equal to 1. A minus I, that means subtract 1 from the diagonal entries of matrix A. So, 1 minus 1, 0, 1 minus 1, 0, 1 minus 1, 0. So, we have 0, 0, 0, 1, 0, 0 and 1, 1, 0. The corresponding equation, the first equation is x is equal to 0 and the second equation from the third row x plus y equal to 0 and we don't have any value or any equation for z therefore z belongs to r. So x is equal to 0 then y is also 0. So we have x 0 y 0 and z belongs to r and the corresponding solution capital x is equal to x y z 0 0 z so z into 0 0 1. So therefore eigenvector is 0 0 1 that means we have only one eigenvector corresponding to lambda equal to 1. Therefore the geometric multiplicity of lambda is number of eigenvector associated to the lambda and we have only one eigenvector associated the lambda equal to 1. So geometric multiplicity is 1. We already see the algebraic multiplicity is 3. Therefore they are not equal A is not diagonalizable. So the first option is wrong. Now for the second option if the minimal polynomial of a is x minus 1 power 3 then a is diagonalizable. So we have one result m is diagonalizable if and only if its minimal polynomial is a product of distinct monic linear factors. Here the minimal polynomial x minus 1 power 3 which is not a product of distinct linear factors Therefore, a is not diagonalizable from this result and hence the second option is also wrong. Now, the third option, the characteristic polynomial of a square is x minus 1 power 3. We are going to use some result. If lambda is an eigenvalue of A, then lambda square is an eigenvalue of A square. Therefore, the characteristic polynomial of A is x minus 1 power 3 is given. Therefore, the eigenvalues of A are 1, 1, 1. And hence the eigenvalue for A square is also 1 square, 1 square, 1 square which is 1, 1, 1 and the second result the eigenvalues are the roots of the characteristic equation. So the characteristic equation for A square is x minus 1 x minus 1 x minus 1 x minus 1 1 power 3 equal to 0 and the characteristic polynomial for A square is x minus 1 power 3 and the option 3 is correct. Now, option 4. Exactly two Jordan blocks, then A minus I power 2 is diagonalizable. So, A has exactly two Jordan blocks. So, A is in the form where two possibilities A equal to 1 1 0, 0 1 0 and 0 0 1 or 1 0 0, 0 1 1, 0 0 1. So, these are the two Jordan blocks 1 1 0 1 and 1 1 1 1 0 1. The order of square matrix is 3 with the single eigenvalue 1. So the two Jordan blocks are of the order 2 plus 1 equal to 3 or 1 plus 2 equal to 3. Therefore we have these two different Jordan blocks. First is of order 2 and the second is of order 1. So for more details click on the suggested video top of the screen. Now we required A minus I whole square. So first we calculate the A minus I. To subtract One from the diagonal entries of matrix A, so it is 1, minus 1, 0, 0, 0. So, A minus I is 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0. And for the second option, A minus I, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0. In both the case, the square of A minus I will be 0 matrix and the 0 matrix is always diagonalizable. Therefore, the fourth option is correct. Which is, if A has exactly two Jordan blocks, then A minus I whole square is diagonalizable. Hence, this question has two correct answers, third and the fourth. Thanks for watching. For more videos, like, subscribe, press the bell icon of Next Gen Man.