#csirnet #mathematics #linearalgebra FEB 2022 | Dimension of Eigen space and Diagonalization

Hello friends, this is Mahesh Yuvalekar and you are watching NextGenMaths. If you have not subscribed my channel, then please subscribe the channel and press the bell icon. In this video, we are going to discuss the problem of CSI NET FEB 2022 problem of Linear Algebra. The question is, let M equal to 3x3 matrix 0-1 0, 1 2-1 1 1 3. Given that 1 is the eigenvalue of M, which of the following statements is true? We are given four options. First option is, minus 2 is an eigenvalue of M, second option is 3 is an eigenvalue of M, third option, the eigenspace of each eigenvalue has dimension 1 and the fourth option M is diagonalizable matrix. Solution, the first two option is about the eigenvalue of M and the last two option is about the eigenvector of M. So, here we are going to find the eigenvalue and eigenvector of matrix M. The given matrix M is 0-1 0, 1 2-1 1 1 3. We have also given one of the eigenvalue of matrix M is 1. M is 3x3 matrix, so it has three eigenvalues. Suppose lambda 1, lambda 2 and lambda 3. But one of the eigenvalue is given which is 1, so lambda 1 equal to 1. Now we try to find the other two eigenvalues lambda 2 and lambda 3. For that, we are going to use some property of eigenvalues. So first property of eigenvalues, product of eigenvalue is equal to determinant of matrix. So lambda 1 into lambda 2 into lambda 3 equal to determinant of M. Lambda 1 equal to 1, so 1 into lambda 2 into lambda 3 and determinant value of M is minus 1 into minus 1, 3 plus 1 which will give the 4. So lambda 2 into lambda 3 equal to 4. Now another property of eigenvalue is sum of eigenvalue equal to trace of matrix. Lambda 1 plus lambda 2 plus lambda 3 equal to trace of M. Lambda 1 equal to 1, so 1 plus lambda 2 plus lambda 3 equal to trace of M is sum of its diagonal value which is 0 plus 2 plus 3 equal to 5. Lambda 1 plus lambda 2 plus lambda 3 equal to 5, so lambda 2 plus lambda 3 equal to 4. Now, we have two equations lambda 2 into lambda 3 equal to 4 and lambda 2 plus lambda 3 equal to 4. We try to find the lambda 2 and lambda 3 from these two equations. So, for that we calculate square of lambda 2 minus lambda 3 equal to lambda 2 plus lambda 3 whole square minus 4 times lambda 2 into lambda 3. Lambda 2 plus 3 equal to 4 and lambda 2 into 3 equal to 4. So, 4 square minus 4 into 4 will give the 0. Therefore, lambda 2 minus lambda 3 whole square equal to 0 which will give lambda 2 equal to lambda 3. So, lambda 2 equal to lambda 3, lambda 2 into lambda 3 equal to 4 and lambda 2 plus lambda 3 equal to 4. Therefore, it will give the lambda 2 equal to lambda 3 equal to 2. Hence, 1, 2, 2 are the eigenvalues of matrix M. But option 1 says that minus 2 is an eigenvalue of M which is wrong and option 2 says that 3 is an eigenvalue of M which is also false. So, first and second option are the wrong option. Now, for the next two option we need the eigenvectors of matrix M. 1, 2, 2 are the eigenvalues of matrix M. Here, eigenvalue 1 occur only once. Therefore, there is only one eigenvector associated with the eigenvalue 1. And hence, the eigenspace of an eigenvalue 1 has dimension 1. So, we are not going to find the eigenvector associated to the 1 but its eigenspace has dimension 1 because it contain only one independent eigenvector as eigenvalue occurred only once. Now, the second eigenvalue 2 is occurred twice. So, we need to find its eigenvector. We cannot predict how many eigenvectors associated with the 2. The eigenvectors of eigenvalue 2 are the solution of system of linear equation M minus 2i into x equal to 0. So, M minus 2i, here i is an identity matrix of order 3 cross 3. So, M minus 2i can be obtained by subtracting minus 2 from the diagonal entries of matrix M. So, 0 minus 2 is minus 2, 2 minus 2 is 0 and 3 minus 2 is 1 and other values are the same into x, y, z equal to 0, 0, 0. To solve the system of linear equation, we are going to use the Gauss order elimination method. So, a augmented with the 0, minus 2, minus 1, 0, 0, 1, 0, minus 1, 0, 1, 1, 1, 0. Now, apply the row operation and convert into the reduced row echelon form. Then we have 1, 0, minus 1, 0, 1, 2 and 0, 0, 0. It will give the equation x minus z equal to 0, y plus 2z equal to 0. Therefore, we have x equal to z and y equal to minus 2z. Therefore, the solution capital X equal to x, y, z, x equal to z, y equal to minus 2z and z equal to z. So, z into 1 minus 2, 1. So, 1 minus 2, 1 is an eigenvector associated with the eigenvalue lambda equal to 2. Therefore, eigenspace of eigenvalue 2 equal to z into 1 minus 2, 1, z belongs to R. There is only one eigenvector in the eigenspace of eigenvalue 2. Therefore, the dimension of eigenspace of eigenvalue 2 is 1. Hence, the dimension of eigenspace of both the eigenvalue is 1. Therefore, the option number 3 is the correct option which is the eigenspace of each eigenvalue as dimension 1. Now, for the option number 4, algebraic multiplicity of eigenvalue 2 is 2 and geometric multiplicity of eigenvalue 2 is 1. Therefore, algebraic multiplicity and geometric multiplicity are not equal for the eigenvalue 2. Hence, matrix M is not diagonalizable because matrix M is diagonalizable only when the algebraic multiplicity and geometric multiplicity are equal for all eigenvalues. In our case, they are not equal for the eigenvalue 2. Therefore, M is not a diagonalizable matrix. So, option number 4 is the wrong option. Option number 3 is only the correct option. Thanks for watching. For more videos, like, share, subscribe, press the bell icon of NextGenMaths.