CSIR NET MATHEMATICS| Linear Algebra | How to identify Minimal polynomial from the given option?

In this video, we are going to discuss example of minimal polynomial from CSIR NET December 2018. Example 65 from booklet 3 which is a problem of marks 4.75. Let M be a 3x3 matrix 1-1, 1, 2, 1, 4, –2, 1, –4. Given that 1 is an eigenvalue of A, then which among the following are correct? The first option, the minimal polynomial of M is x minus 1 into x plus 4. Second option, the minimal polynomial of M is x minus 1 square into x plus 4. Third option, M is not diagonalizable. And the fourth option, M inverse is equal to 1 by 4 M plus 3i. Now, we begin with the sum results of eigenvalues. And the first result is. Sum of eigenvalue is trace of A. Suppose lambda 1, lambda 2, lambda n are the eigenvalues of A. Then its sum lambda 1 plus lambda 2 plus lambda n equal to trace of A. And the second result product of eigenvalues of A is determinant to trace of A. That means lambda 1 into lambda 2 into lambda n equal to determinant of A. Now for the given problem M is 3 cross 3 matrix. Therefore We assume lambda 1, lambda 2, and lambda 3 are the eigenvalues of M. So, first result, lambda 1 plus lambda 2 plus lambda 3 equal to trace of M. Trace of M means sum of diagonals. So, 1 plus 1 minus 4 equal to minus 2. And one of the eigenvalues is 1. Suppose lambda 1 equal to 1. So, 1 plus lambda 2 plus lambda 3 equal to minus 2. So, lambda 2 plus lambda 3 equal to minus 3. Now, the second result. Product of eigenvalues lambda 1 into lambda 2 into lambda 3 equal to determinant of A. So lambda 1 equal to 1, lambda 2 into lambda 3 equal to determinant 1 into minus 4 minus 4 plus 1 into minus 8 minus 8 plus 1 into 2 plus 2. So finally the determinant has value minus 4. So we have second equation lambda 2 into lambda 3 equal to minus 4. By simple elimination Between this two equation, we can find lambda 2 equal to 1 and lambda 3 equal to minus 4. You can see 1 minus 4 equal to minus 3 and 1 into minus 4 equal to minus 4. Therefore, lambda 1 equal to 1, lambda 2 equal to 1 and lambda 3 equal to minus 4 are the eigenvalues of A. Therefore, the characteristic equation is p of x is equal to x minus 1, x minus 1, x plus 4 because lambda 1, lambda 2, lambda 3 are the roots of characteristic equation and the final form is x minus 1 whole square into x plus 4 equal to 0. Definition of minimal polynomial. The minimal polynomial m(x) of matrix M is a monic polynomial of smallest degree such that m(M) equal to 0. Note that The minimal polynomial Always divide characteristic equation For the option one. Suppose x minus 1, x plus 4 is minimal polynomial, then matrix M satisfies minimal polynomial. That means m(M), M minus I, M plus 4I should be 0. So, M minus I, that means subtract 1 from the diagonal entry, so 0, 0 and minus 5. And M plus 4I, that means add 4 into the diagonal entries. So, 1 plus 4 is 5, 1 plus 4 is 5 and minus 4 plus 4 equal to. Now, we are going to calculate product. If product has value 0, then the first option is correct. That is x minus 1, x plus 4 is a minimal polynomial. We calculate the first value. Minus 1, minus 2, so minus 4, so first value is non-zero, that means for matrix is non-zero, we are not going to calculate the other values and we can see that the first option is wrong, therefore x minus 1, x plus 4 is not a minimal polynomial. Now, for the second option, m(x) is equal to x minus 1 square, x plus 4, which is nothing but the characteristic polynomial. So m(M). equal to phi of M equal to zero because every matrix satisfy characteristic equation Cayley Hamilton theorem therefore phi of M equal to zero and therefore the second option is correct. Now option three we have one result M is diagonalizable if and only if its minimal polynomial is a product of distinct monic linear factors. Our minimal polynomial is x minus 1 square x plus 4 which is not a product of distinct factors. Therefore, m is not a diagonalizable matrix and hence the third option is correct. Now, for the fourth option, the characteristic polynomial of m is phi of x is equal to x minus 1 whole square x plus 4. By the Cayley-Hamilton theorem, phi of m equal to zero because matrix M satisfies characteristic equation and M minus i whole square M plus 4i equal to 0. If we open the brackets and convert into M inverse equal to 1 by 4 minus M square minus 2M plus 7i which is not the fourth option. So, fourth option is wrong. So, we have only two correct option. Thanks for watching. For more videos like, subscribe, please the bell icon of next gen.