CSIR NET MATHEMATICS| Linear Algebra | How to Calculate Jordan Canonical Form? (June 2018)

In this video, we are going to discuss Jordan Canonical Form from CSINET, June 2018. Example number 75, booklet A, marks 4.75. Example, let A be 6 x 6 matrix over R with characteristic polynomial x-3 whole square x-2 power 4 and minimal polynomial (x-3)(x-2)^2. Then Jordan canonical form of A can be the first matrix, the diagonals are 3, 3, 2, 2, 2, 2 and in the upper off-diagonal entry is 1, 1, 1. The second option, diagonal 3, 3 and 4, 2 and in the second off-diagonal 1 and similar for the other two options. Now, answer. First, we are going to understand the Jordan canonical form of any matrix A. So, if matrix A of order n cross n then the Jordan canonical form of matrix A is also of order n cross n and expresses matrix J1, J2, Jk where J1, J2, Jk are Jordan blocks of the form. J1 equal to the diagonals are the lambda 1, lambda 1, lambda 1 are the same and off diagonals are 1, 1, 1. lambda 2 lambda 2 of diagonal is 1 similarly j3 lambda 3 lambda 3 lambda 3 of diagonal is 1 and jk lambda k where lambda 1 lambda 2 lambda k are the eigenvalues of k. That means if we have eigenvalues of lambda 1 lambda 2 lambda k then corresponding we have Jordan blocks and based on its algebraic multiplicity we are able to find this Jordan blocks. Now How can we calculate Jordan canonical form using characteristic polynomial and minimal polynomial? It is compared to easy direct finding Jordan canonical form if we have given characteristic polynomial and minimal polynomial. Suppose the characteristic polynomial of any matrix is x minus lambda 1 power a1, x minus lambda 2 power a2, x minus lambda r power ar. So lambda 1, lambda Lambda to Lambda RRD. eigenvalues. a1, a2, ar are the algebraic multiplicities and corresponding minimal polynomial is x minus lambda 1 power c1, x minus lambda 2 power c2, x minus lambda r power cr. Always remember c1 less than equal to a1, c2 less than equal to a2 and cr less than equal to ar. Then the order of Jordan block for lambda i D1, D2, DK with the condition D1 plus D2 plus DK is equal to AI and highest order of Jordan block is ci. That means for particular lambda 1, the order of Jordan blocks for lambda equal to 1 are D1, D2, DK with the condition the highest value of D1, D2, DK is c1. The sum of order should be a1. So c1 a1 is a power of x minus lambda in minimal polynomial and a1 is the algebraic multiplicity of lambda. Now, in our case, we have given characteristic polynomial x minus 3 power 2 and x minus 2 power 4. Therefore, algebraic multiplicity for lambda equal to 3 is 2 and algebraic multiplicity of lambda equal to 2 is 4. And the minimal polynomial is x minus 3 power 1 and x minus 2 power 2. Therefore, c1 equal to 1 and c2 equal to 2, A1 equal to 2 and A2 equal to 4. Now, for lambda 1 equal to 3, highest power of Jordan block is C1 equal to 1. The Jordan blocks are in the form D1 equal to 1, D2 equal to 1 because D1 plus D2 is equal to 2. This is A1. Therefore, the Jordan blocks for lambda 1 equal to 3 are J1 equal to 3 and J2 is equal to 3 because we have not any other possibility because highest order of Jordan block is C1 equal to 1. So all the blocks are of order 1 and therefore 1 plus 1 the sum is now. For lambda equal to 2 the highest order Jordan block is C2 equal to 2. So other blocks may be of order 1 or may be of order 2. So, lambda 2 equal to 2, so if we take other Jordan block of order 2, so highest order is 2, you can take any other order 2 or 1, so if we take another order, other Jordan block of order 2, so d1 plus d2 is equal to 4, but that is also possibility the Jordan block is of order 1, then d1 equal to 2, the highest order and the other Jordan block of order 1 and 1, so total sum is D1 plus D2 plus D3 is equal to 2 plus 1 plus 1 equal to 4. So, we have two different possibilities for the Jordan block. The first possibility is the highest order should be 2. So, we can take D1 equal to 2 and D2 equal to 2. So, D1 plus D2 equal to 4, but highest order is 2. So, lower order may be possible, 1, 1. So, that's why D1 equal to 2, D2 equal to 1 and D3 is equal to 1. So, 2 plus 1 plus 1 equal to 4. So Jordan blocks for lambda equal to 2 has two different possibilities. First possibility, J3 is equal to of order 2, therefore 2 cross 2 matrix, 2, 1, 0, 2. J4 is equal to 2, 1, 0, 2. The first possibility, 2 plus 2. And the another possibility, the highest order is 2, so one block must be of order 2. So J3 is equal to 2, 1, 0, 2. And the other blocks of order 1, so J4 equal to 2 and J5 equal to 2. Therefore, the final canonical form of matrix A has two possibilities J1, J2, J3, J4 and other possibilities J1, J2, J3, J4, J5. So, these are the calculated J1, J2, J3, J4, J5 and because of that the corresponding Jordan canonical form of matrix A is 3, 3, the J1, J2 then J3, 2, 2, 1. and J4 is also of order 2, so 2, 2, 1 and another possibility J1, J2, J3, J4, J5, so J1 is 3, J2 is 3, J3 is 2, 2, 1 and J4 is 2 and J5 is 2, so therefore the correct options are number 2 which is of the form J1, J2, J3, J4 and J5, this one and third option Which is of the form J1, J2, J3, J4. So, 2 and 3 are the correct option. Thanks for watching. For more videos, like, subscribe, raise the bell icon of next gen.