CSIR NET MATHEMATICS | Linear Algebra | Jordan Canonical Form DEC 2011

Welcome friends, in this video we are going to discuss the problem of linear algebra from CSI in December 2011. Let matrix A equal to 0 0 minus 4 1 0 0 0 1 0 5 0 0 1 0. Then the Jordan canonical form of A is, we have 4 options, 4 different matrices. Now to find the Jordan canonical form of matrix A, we are required to find characteristic polynomial of A. So characteristic polynomial is determinant lambda i minus a and if we evaluate this determinant then we have lambda power 4 minus 5 lambda square plus 4. Now the factors of this polynomial is lambda square minus 4 lambda square minus 1 and further factor lambda minus 2 lambda plus 2 lambda minus 1 lambda plus 1. Now eigenvalues are the root of the characteristic equation determine lambda i minus a equal to 0 so this is the characteristic polynomial equate with the 0 then we have four roots lambda equal to 2 lambda equal to minus 2 lambda equal to 1 and lambda equal to minus 1 these are the eigenvalues of matrix a all the eigenvalues are distinct therefore a is diagonalizable matrix and for the diagonalizable matrix the jordan canonical form is nothing but the diagonal form of the matrix therefore the diagonal form of the matrix is nothing but the diagonal are the eigenvalues and other values are the 0 so diagonal values are the eigenvalues minus 1 1 2 and minus 2 and other values are the 0 which will be the jordan canonical form for the given matrix and which is a option number 1 is a correct answer thanks for watching for more videos like share subscribe press the bell icon of next gen yes