Question

no calculator please

1 (8 pts) Find the dimension and a basis for the following vector spaces. (a) (4 pts) The vector space of all symmetric 2 x 2 matrices (which is a subspace of M22). (b) (4 pts) All vectors of the form (a, b, 2a + 3b) (which is a subspace of R®). 2. (12 pts) Given the matrix in a R R-E form: 1000 3 0110-2 00011 0 0 0 0 0 (a) (6 pts) Find rank(A) and mullity(A), and mullity(A”). 1 1 0 (b) (2 pts) Find a basis for the row space of A. (c) (2 pts) Find a basis for the column space of A. (d) (2 pts) Find a basis for the null space of A.

Answer 1

3 solution 1 : (a) Vector space of all cymmetric 2x2 matrices is given by { [0 b] la, b, c E IF}; where it is a field. Dimension Basis {[68], [0],[]} (6) All vectors of the form (a, b, 2a +36) can be written as { (a, b, 2a + 3b) (a, b E IF}. Since (a, b, 29+3b) = a (1,0,2) + b(0,1,3) :.{(a, b, 20+3b)} = Linear span of {(1,0,2), (0,1,3). : Dimension = 2 Basis = {(1,0,2), {(1,0,2), (0, 1,3)}. O - 2 solution 2) The matrix A= O 3 0-2 0 O 0 O O are (Another) Since, (method (a) & Since, the last row consists of os and 1st, 2nd and fourth columns ernearly independent, hence rank (A) = 3. [Note: Ronk (A) can't be 4 due to last row] the given matrix is in Row Reduced Echelon form, the number of is the rank of matrix. Therefore rank (A) = 3. Nullity of A 6-3= 3. Explanation. The ronk of a matrix A plus the nullity of A equals the number of columns of A. non-zeno YOWS *

4-3=1. Nullity (AT) Explanation since rank (A)=rank (AT) = 3 and the sumber of columns in A is 4. Hence, using the result mentioned in solution of Nullity (A), we have Nullity (AT) - 4-3=1 (6) Since the matrix is already in Row Reduced Echelon form, the non-zero rows forms. a basis for the Yow space Hence the basis is {(1,0,0,0,3,-2), (0,1,1,0-2, 1), (0,0,0,1,1,1)}. (c) Since, 'column Jank (A) ronk (A) =3. Therefore, it is sufficient to fmd three lineasly independent columns of A. clearly, the 1st, 2nd and 4th re columns are linearly independent. Hence basis for column space is {[:],[8] []} 2 (d) To find the null space, consider the equation 3 - 2 OOO 12 - 2 O-OO OO- O which yields OOOOOO 14 0 O Xς the equations x, +38- 2x=0 x = -3x + 2x6 X+ Sz - 2x + x = 0 x = -x₂ + 2x - X6 + x + x = 0 => x 4 = -x5-x6 + 14

(x,,X2, X₂, 34, X5, X6) = (-3X6 + 2X6, -X2 + 2X-X6, X3, 7 x Ds, X6) = (0,-1,1,0,0,0)+36 (-3,2,0,-),1,0) + < (2,-1, 0,-1,0,1), Hence, the basis for null space is given by 2