please sir help me State the reason briefly for T and give a

please sir help me

State the reason briefly for \"T\" and give a counter-example for \"F\". (1) Let F be a vector space and U, W be subspaces of V. Then U W is a subspace of V . (2) Let V be a vector space and U, W be subspaces of V. Then U W is a subspace of V . (3) Suppose that a matrix (AB) R^ntimesn is nonsingular. Then we can say that (AB)^-1 = B^-1 A^-1 . (4) Given a matrix A^mtimesn , its row-echelon form is unique. (5) Let S= {t^2+ 1, 2t, t^2 + 2t + 1 }. Then S spans a space of polynomials of degree less than or equal to 2. (6) Suppose that A and B are invertible matrices. Then A + B is invertible. (7) Consider the set of real numbers (x, y) with the operations {x_1, y_1) + (x_2 + y_2) = (x_1 + x_2 , y_1 + y_2) and c (x, y) = (cx, y) . Then this set is a vector space. (8) Let A R^ntimesn be skew-symmetric and invertible. Then A^-1 is also skew-symmetric.

Solution

“T” The intersection of two subspaces is closed under addition and scalar multiplication and is, therefore, a subspace. “F” For example, take U to be the x-axis and V the y-axis, both subspaces of R2. Their union includes both (1,0) and (0,1), whose sum, (1,1), is not in the union. Hence, the union is not a vector space. “T” (AB)( B-1 A-1) = A ( BB-1)A-1 = A I A-1 = AA-1 = I. Therefore, (AB)-1 = B-1 A-1 “T” It is a well known algebraic fact. The proof is quite long. “F” Any arbitrary 2nd degree polynomial is of the form at2 + bt + c. If a , b and c are not multiples of 2, then at2 + bt + c is not a linear combination of t2 + 1, 2t and t2 + 2t + 1. “F” Let A = In and B = - In . Both A and B are invertible, but A + B = 0 which is not invertible. “T” since the sum of 2 real numbers and the scalar multiple of a real number are real numbers, the given set is closed under addition and scalar multiplication and is , therefore, a vector space. “T” A skew symmetric matrix satisfies the condition AT = - A. ( A-1)T = (AT)-1 = ( -A )-1 = -A-1