CSIR NET MATHEMATICS DECEMBER 2018 | Linear Algebra | Solutions of Section C (Part 2)

Exam from the net exam December 2018, which is a problem of linearity. Let P3 be the vector space of polynomials with real coefficients and of the degree at most 3. Consider the linear map T from P3 to P3 defined by T of Px is equal to Px plus 1 plus Px minus 1. Which of the following property does the matrix of T with respect to the standard basis be 1x x square x cube of P3 satisfy? The options are, first option, determinant of t equal to 0. The second option, t minus 2i power 4 equal to 0, but t minus 2i power 3 not equal to 0. Third option, t minus 2i power 3 equal to 0, but t minus 2i power 2 not equal to 0. And the fourth option, t is an eigenvalue with multiplicity. The given linear transformation, t of px is equal to p of x plus 1 plus p of x minus 1. Now we are going to find them. Matrix associated to given transformation, for that we require images of basis vector with respect to t. So, first basis vector, p of x is equal to 1, then p of x plus 1 equal to 1, p of x minus 1 is also 1. Then for corresponding image, t of 1 equal to 1 plus 1 equal to 2, in the form of cubic polynomial, it is 2 plus 0 into x plus 0 into x square plus 0 into x cube. Then the column vector form of images. 2, 0, 0, 0. Now, the second basis vector, p of x is equal to x. So, p of x plus 1 equal to x plus 1 and p of x minus 1 is equal to x minus 1. And t of x is equal to sum of p of x plus 1, x minus 1. So, it is x plus 1 plus x minus 1 is equal to 2x. In the cubic polynomial, it is 0 plus 2x plus 0x square plus 0x cube. Then the column vector form of image is 0, 2, 0, 0. Now the third basis vector p of x is equal to x square. So p of x plus 1 equal to x plus 1 whole square and p of x minus 1 is equal to x minus 1 whole. Therefore corresponding image is x plus 1 whole square plus x minus 1 whole square. By simplifying the expression we have 2 plus 2x square. So in the cubic polynomial it is 2 plus 0x plus 2x square plus 0x cube. Therefore, the corresponding column vector 2, 0, 2, 0 and the fourth basis vector p of x is equal to x cube. So, p of x plus 1 equal to x plus 1 whole cube and p of x minus 1 is equal to x minus 1. And the corresponding image with respect to t is x plus 1 whole cube plus x minus 1 whole cube. Now, simplifying the expression, we have 2x cube plus 6x is equal to 0 plus 6x plus 0x square plus 2x cube. Therefore, the corresponding column vector is 0, 6, 0, 2. Now, the T matrix associated with the given transformation T is equal to 1st column T1, 2nd column TX, 3rd column TX square and the 4th column TX cube. Subtracting the value of these vectors, we have matrix 2, 0, 0, 0. The 2nd column 0, 2, 0, 0 and the 3rd column 2, 0, 2, 0. Vote column 06. Now, the determinant of t is equal to 16, which is not equal to 0. We have given the option first one, determinant of t equal to 0, which is wrong because the determinant value is 16. The first option is cancelled. Now, the characteristic polynomial of t is determinant of t minus lambda i. Then, we have 2 minus lambda power 4. OOF DAD OOF Two easy. Eigen value with the multiplicity 4, which is a fourth option. So, fourth option is correct. Now, in the second and third option, we require t minus 2i. So, we are going to find first t minus 2i, that means subtract 2 from the diagonal entries of t. So, if we subtract 2 from the diagonal entries of t, then all the diagonals will be 0. So, we have t minus 2i is equal to 0 0 2 0 0 0 0 6 and the remaining two rows are 0. Now we require the 2 3 and the fourth power of t minus 2i. The second power of t minus 2i that means t minus 2i square is equal to 0. Therefore all the remaining powers are 0. So the second and third option are wrong because in the second option T minus 2i power 3 not equal to 0 and the third option T minus 2i power 2 not equal to 0. Therefore, second and third option are cancelled. Only the fourth option is correct which is 2 is an eigenvalue with multiple. Example number 69 from the net exam December 2018 which is a problem of linear algebra. Example, define real valued function B on R2 cross R2 as follows. If v equal to x1, x2 W is equal to y1, y2, belongs to R2. Define, B, V, W is equal to x1, y1, minus x1, y2, minus x2, y1, plus 4x2, y2. Let V0 is equal to 1, 0. And, W equal to, V belongs to R2. B, V, 0, V equal to 0. Then, first option, W is not subspace of R2. Second option, W equals 0, 0. Third option. W is the y-axis and the fourth option, W is the line passing through origin and 1-1. Answer, we have given V equal to x1, x2. W is equal to y1, y2. Then the function B is defined as BVW equal to x1, y1 minus x1, y2 minus x2, y1 plus 4x2, y2. V0 is equal to 1, 0. W is equal to W belongs to R2. BV0, V equal to 0. The set W is defined as bv0 v equal to 0. If we consider as a y1 y2 belongs to R2 then b1 0 y1 y2 equal to 0. In the definition of b we substitute the value 1 0 and y1 y2 then we have we consider x1 is equal to 1 and x2 is equal to 0. So 1 into y1 minus 1 into y2 minus 0 into y1 plus 4 into 0 into We have simplified form. y1 minus y2 equal to 0. So w is a y1 y2 belongs to R2 such that y1 minus y2 equal to 0. So it is a line passing through the O region and as we know every line passing through the O region is a subspace of R2. Therefore the first option that w is not a subspace of R2 which is wrong because w is a subspace of R2. Now the second option, w equals 0, 0. That is not the case because we have other values 1, 1, 2, 2, 3, 3, 4, 4 are also the elements. So second option is also wrong. Now third option is not because w is not in y axis. And the fourth option, w is the line passing through origin 0, 0 and 1, 1 which is correct because 1 minus 1 is satisfied this equation. So fourth option is the final answer. Example number 17, from net exam December 2018, which is a problem of linear algebra. Example, consider the quadratic forms, q1xy equal to xy, q2xy is equal to x square plus 2xy plus y square, and q3xy is equal to x square plus 3xy plus 2y square. On R2, choose the correct statement from the following. The first statement, q1 and q2 are equivalent, q1 and q3 are equivalent, q2 and q3 are equivalent, all are equal. To solve this example, we are required to understand the equivalence of quadratic forms. So, we begin with the definition of equivalence quadratic form. Two quadratic forms in x-y, that is binary quadratic forms, are said to be equivalent if there exists integer a, b, c, d such that f of x-y is equal to g of a x plus b y, c x plus d y and the determinant value of a, b, c, d, a, d minus b, c should be 1. Now we are given the quadratic form q1 xy is equal to xy, q2 xy is equal to x square plus 2xy plus y square which can be written in the form of x plus y whole square and q3 xy is equal to x square plus 3xy plus 2y square which is factorized into x plus 2y x plus y. Now first we are going to check the equivalence of q1 and q2. q1 xy equal to q2 ax plus by comma cx plus by. If we are able to find the integer value of ABCD, Then, we can say that Q1 and Q2 are equivalent. So, X, Y. Now, Q2, X, Y is equal to X plus Y whole square. So, here we replace X by AX plus BY and Y replaced by CX plus DY. So, X, Y is equal to AX plus BY plus CX plus DY whole square. Now, gathering the terms of X and Y, then we have X, Y is equal to A plus CX plus B plus DY whole square. Now, opening the brackets. xy is equal to a plus c whole square x square plus a plus c b plus d xy plus b plus d whole square y square. Now comparing the coefficient of x square xy and y square we have a plus c whole square equal to 0 b plus d whole square equal to 0 a plus c into b plus d equal to 1. That means a plus c equal to 0 b plus d equal to 0 and a plus c into b plus d equal to 1. Which is not possible because a plus c equal to 0, b plus d equal to 0. Then its multiplication would not be 1. So which is not possible. Therefore, the q1 is not equivalent to the q2. The first option is wrong. And the also fourth option is wrong. Which is all are equivalent. We are going to check the equivalence of q2 and q3. So now we assume q2ax plus by, cx plus dy equal to q3xy. Now, ax plus by plus cx plus dy whole square is equal to x square plus 3xy plus 2y square. Now, opening the brackets, a plus c whole square x square plus a plus c into b plus dxy plus b plus d whole square y square equal to x square plus 3xy plus 2y square. By comparing the coefficient of x square, xy and y square, we have this relation a plus c whole square. Equal to 1, b plus d whole square equal to 2 and a plus c into b plus d is equal to 3. But b plus d whole square is equal to 2 which is not possible because a, b, c, d are integers. So, square of any integer should not be 2. So, therefore, this equivalence is also not possible. So, option number 3 is also wrong. Now, we are going to check the equivalence of q1 and q3. q1 x ax plus by and cx plus dy equal to q3 xy. So, AX plus BY into CX plus DY as a definition of Q1XY, X into Y. So, X replaced by AX plus BY and Y replaced by CX plus DY. So, we have AX plus BY into CX plus DY is equal to X square plus 3XY plus 2Y square. Now, factorize the right hand side term. So, we have X plus Y into X plus 2Y. By natural comparison, We can say that one of the value of ABCD is A equal to 1, B equal to 1, C equal to 1 and D equal to 2. AD minus BC is equal to 1. So, ABCD are integers and AD minus BC determinant value is 1. Therefore, we are able to find ABCD such that Q1 of AX plus BY, CX plus DY equal to Q3XY. Therefore, Q1 equivalent to Q3. So, this is the correct answer 2. Thanks for watching. For more videos, like, subscribe, press the bell icon of NextGen.