Section 3.6. [tx, ty] = t[x, y] S.

Transcription

1 Section 3.6. (a) Let S be the set of vectors [x, y] satisfying x = 2y. Then S is a vector subspace of R 2. For (i) [, ] S as x = 2y holds with x = and y =. (ii) S is closed under addition. For let [x, y ] and [x 2, y 2 ] belong to S. Then x = 2y and x 2 = 2y 2. Hence x + x 2 = 2y + 2y 2 = 2(y + y 2 ) and hence belongs to S. [x + x 2, y + y 2 ] = [x, y ] + [x 2, y 2 ] (iii) S is closed under scalar multiplication. For let [x, y] S and t R. Then x = 2y and hence tx = 2(ty). Consequently [tx, ty] = t[x, y] S. (b) Let S be the set of vectors [x, y] satisfying x = 2y and 2x = y. Then S is a subspace of R 2. This can be proved in the same way as (a), or alternatively we see that x = 2y and 2x = y imply x = 4x and hence x = = y. Hence S = {[, ]}, the set consisting of the zero vector. This is always a subspace. (c) Let S be the set of vectors [x, y] satisfying x = 2y +. Then S doesn t contain the zero vector and consequently fails to be a vector subspace. (d) Let S be the set of vectors [x, y] satisfying xy =. Then S is not closed under addition of vectors. For example [, ] S and [, ] S, but [, ] + [, ] = [, ] S. (e) Let S be the set of vectors [x, y] satisfying x and y. Then S is not closed under scalar multiplication. For example [, ] S and R, but ( )[, ] = [, ] S. 2. Let X, Y, Z be vectors in R n. Then by Lemma 3.2. X + Y, X + Z, Y + Z X, Y, Z, as each of X + Y, X + Z, Y + Z is a linear combination of X, Y, Z. 3

2 Also X = 2 (X + Y ) + 2 (X + Z) (Y + Z), 2 so Hence 3. Let X = Y = 2 (X + Y ) 2 (X + Z) + (Y + Z), 2 Z = 2 (X + Y ) + 2 (X + Z) + (Y + Z), 2 2 X, Y, Z X + Y, X + Z, Y + Z. X, Y, Z = X + Y, X + Z, Y + Z., X 2 = 2 and X 3 = 3. We have to decide if X, X 2, X 3 are linearly independent, that is if the equation xx + yx 2 + zx 3 = has only the trivial solution. This equation is equivalent to the folowing homogeneous system x + y + z = x + y + z = x + y + z = 2x + 2y + 3z =. We reduce the coefficient matrix to reduced row echelon form: and consequently the system has only the trivial solution x =, y =, z =. Hence the given vectors are linearly independent. 4. The vectors X = λ, X 2 = λ, X 3 = λ 3

3 are linearly dependent for precisely those values of λ for which the equation xx +yx 2 +zx 3 = has a non trivial solution. This equation is equivalent to the system of homogeneous equations λx y z = x + λy z = x y + λz =. Now the coefficient determinant of this system is λ λ λ = (λ + )2 (λ 2). So the values of λ which make X, X 2, X 3 linearly independent are those λ satisfying λ and λ Let A be the following matrix of rationals: 2 A = Then A has reduced row echelon form B = From B we read off the following: 3 (a) The rows of B form a basis for R(A). (Consequently the rows of A also form a basis for R(A).). (b) The first four columns of A form a basis for C(A). (c) To find a basis for N(A), we solve AX = and equivalently BX =. From B we see that the solution is x = x 5 x 2 = x 3 = x 5 x 4 = 3x 5, 32

4 with x 5 arbitrary. Then X = x 5 x 5 3x 5 x 5 = x 5 so [,,, 3, ] t is a basis for N(A). 3, 6. In Section.6, problem 2, we found that the matrix A = has reduced row echelon form B = From B we read off the following:. (a) The three non zero rows of B form a basis for R(A). (b) The first three columns of A form a basis for C(A). (c) To find a basis for N(A), we solve AX = and equivalently BX =. From B we see that the solution is x = x 4 x 5 = x 4 + x 5 x 2 = x 4 x 5 = x 4 + x 5 x 3 = x 4 = x 4, with x 4 and x 5 arbitrary elements of Z 2. Hence x 4 + x 5 x 4 + x 5 X = x 4 x 4 = x 4 + x 5 x 5. Hence [,,,, ] t and [,,,, ] t form a basis for N(A). 33

5 7. Let A be the following matrix over Z 5 : A = We find that A has reduced row echelon form B: 2 4 B = From B we read off the following: (a) The four rows of B form a basis for R(A). (Consequently the rows of A also form a basis for R(A). (b) The first four columns of A form a basis for C(A). (c) To find a basis for N(A), we solve AX = and equivalently BX =. From B we see that the solution is x = 2x 5 4x 6 = 3x 5 + x 6 x 2 = 4x 5 4x 6 = x 5 + x 6 x 3 = x 4 = 3x 5 = 2x 5, where x 5 and x 6 are arbitrary elements of Z 5. Hence 3 X = x x 6, so [3,,, 2,, ] t and [,,,,, ] t form a basis for N(A). 8. Let F = {,, a, b} be a field and let A be the following matrix over F: a b a A = a b b. a 34

6 In Section.6, problem 7, we found that A had reduced row echelon form B = b. From B we read off the following: (a) The rows of B form a basis for R(A). (Consequently the rows of A also form a basis for R(A). (b) The first three columns of A form a basis for C(A). (c) To find a basis for N(A), we solve AX = and equivalently BX =. From B we see that the solution is x = x 2 = bx 4 = bx 4 x 3 = x 4 = x 4, where x 4 is an arbitrary element of F. Hence X = x 4 b, so [, b,, ] t is a basis for N(A). 9. Suppose that X,...,X m form a basis for a subspace S. We have to prove that X, X + X 2,...,X + + X m also form a basis for S. First we prove the independence of the family: Suppose Then x X + x 2 (X + X 2 ) + + x m (X + + X m ) =. (x + x x m )X + + x m X m =. Then the linear independence of X,...,X m gives x + x x m =,...,x m =, 35

7 form which we deduce that x =,...,x m =. Secondly we have to prove that every vector of S is expressible as a linear combination of X, X + X 2,...,X + + X m. Suppose X S. Then We have to find x,...,x m such that Then X = a X + + a m X m. X = x X + x 2 (X + X 2 ) + + x m (X + + X m ) = (x + x x m )X + + x m X m. a X + + a m X m = (x + x x m )X + + x m X m. So if we can solve the system x + x x m = a,...,x m = a m, we are finished. Clearly these equations have the unique solution x = a a 2,...,x m = a m a m, x m = a m. [ ] a b c. Let A =. If [a, b, c] is a multiple of [,, ], (that is, a = b = c), then ranka =. For if then [a, b, c] = t[,, ], R(A) = [a, b, c], [,, ] = t[,, ], [,, ] = [,, ], so [,, ] is a basis for R(A). However if [a, b, c] is not a multiple of [,, ], (that is at least two of a, b, c are distinct), then the left to right test shows that [a, b, c] and [,, ] are linearly independent and hence form a basis for R(A). Consequently ranka = 2 in this case.. Let S be a subspace of F n with dims = m. Also suppose that X,...,X m are vectors in S such that S = X,...,X m. We have to prove that X,...,X m form a basis for S; in other words, we must prove that X,...,X m are linearly independent. 36

8 However if X,...,X m were linearly dependent, then one of these vectors would be a linear combination of the remaining vectors. Consequently S would be spanned by m vectors. But there exist a family of m linearly independent vectors in S. Then by Theorem 3.3.2, we would have the contradiction m m. 2. Let [x, y, z] t S. Then x + 2y + 3z =. Hence x = 2y 3z and x y z = 2y 3z y z = y 2 + z Hence [ 2,, ] t and [ 3,, ] t form a basis for S. Next ( ) + 2( ) + 3() =, so [,, ] t S. To find a basis for S which includes [,, ] t, we note that [ 2,, ] t is not a multiple of [,, ] t. Hence we have found a linearly independent family of two vectors in S, a subspace of dimension equal to 2. Consequently these two vectors form a basis for S. 3. Without loss of generality, suppose that X = X 2. Then we have the non trivial dependency relation: X + ( )X 2 + X X m = (a) Suppose that X m+ is a linear combination of X,...,X m. Then X,...,X m, X m+ = X,...,X m and hence dim X,...,X m, X m+ = dim X,...,X m. (b) Suppose that X m+ is not a linear combination of X,...,X m. If not all of X,...,X m are zero, there will be a subfamily X c,...,x cr which is a basis for X,...,X m. Then as X m+ is not a linear combination of X c,...,x cr, it follows that X c,...,x cr, X m+ are linearly independent. Also Consequently X,...,X m, X m+ = X c,...,x cr, X m+. dim X,...,X m, X m+ = r + = dim X,...,X m +. 37

9 Our result can be rephrased in a form suitable for the second part of the problem: dim X,...,X m, X m+ = dim X,...,X m if and only if X m+ is a linear combination of X,...,X m. If X = [x,...,x n ] t, then AX = B is equivalent to B = x A + + x n A n. So AX = B is soluble for X if and only if B is a linear combination of the columns of A, that is B C(A). However by the first part of this question, B C(A) if and only if dimc([a B]) = dimc(a), that is, rank [A B] = rank A. 5. Let a,...,a n be elements of F, not all zero. Let S denote the set of vectors [x,...,x n ] t, where x,...,x n satisfy a x + + a n x n =. Then S = N(A), where A is the row matrix [a,...,a n ]. Now ranka = as A. So by the rank + nullity theorem, noting that the number of columns of A equals n, we have dim N(A) = nullity (A) = n ranka = n. 6. (a) (Proof of Lemma 3.2.) Suppose that each of X,...,X r is a linear combination of Y,...,Y s. Then s X i = a ij Y j, j= ( i r). Now let X = r i= x ix i be a linear combination of X,...,X r. Then X = x (a Y + + a s Y s ) + + x r (a r Y + + a rs Y s ) = y Y + + y s Y s, where y j = a j x + +a rj x r. Hence X is a linear combination of Y,...,Y s. Another way of stating Lemma 3.2. is X,...,X r Y,...,Y s, () 38

10 if each of X,...,X r is a linear combination of Y,...,Y s. (b) (Proof of Theorem 3.2.) Suppose that each of X,...,X r is a linear combination of Y,...,Y s and that each of Y,...,Y s is a linear combination of X,...,X r. Then by (a) equation () above X,...,X r Y,...,Y s and Hence Y,...,Y s X,...,X r. X,...,X r = Y,...,Y s. (c) (Proof of Corollary 3.2.) Suppose that each of Z,...,Z t is a linear combination of X,...,X r. Then each of X,...,X r, Z,...,Z t is a linear combination of X,...,X r. Also each of X,...,X r is a linear combination of X,...,X r, Z,...,Z t, so by Theorem 3.2. X,...,X r, Z,...,Z t = X,...,X r. (d) (Proof of Theorem 3.3.2) Let Y,...,Y s be vectors in X,...,X r and assume that s > r. We have to prove that Y,...,Y s are linearly dependent. So we consider the equation x Y + + x s Y s =. Now Y i = r j= a ijx j, for i s. Hence x Y + + x s Y s = x (a X + + a r X r ) + + x r (a s X + + a sr X r ). = y X + + y r X r, () where y j = a j x + + a sj x s. However the homogeneous system y =,, y r = has a non trivial solution x,...,x s, as s > r and from (), this results in a non trivial solution of the equation x Y + + x s Y s =. 39

11 Hence Y,...,Y s are linearly dependent. 7. Let R and S be subspaces of F n, with R S. We first prove dimr dims. Let X,...,X r be a basis for R. Now by Theorem 3.5.2, because X,...,X r form a linearly independent family lying in S, this family can be extended to a basis X,...,X r,...,x s for S. Then dim S = s r = dimr. Next suppose that dimr = dims. Let X,...,X r be a basis for R. Then because X,...,X r form a linearly independent family in S and S is a subspace whose dimension is r, it follows from Theorem that X,...,X r form a basis for S. Then S = X,...,X r = R. 8. Suppose that R and S are subspaces of F n with the property that R S is also a subspace of F n. We have to prove that R S or S R. We argue by contradiction: Suppose that R S and S R. Then there exist vectors u and v such that u R and u S, v S and v R. Consider the vector u+v. As we are assuming R S is a subspace, R S is closed under addition. Hence u + v R S and so u + v R or u + v S. However if u + v R, then v = (u + v) u R, which is a contradiction; similarly if u + v S. Hence we have derived a contradiction on the asumption that R S and S R. Consequently at least one of these must be false. In other words R S or S R. 9. Let X,...,X r be a basis for S. (i) First let Y = a X + + a r X r. (2) Y r = a r X + + a rr X r, 4

12 where A = [a ij ] is non singular. Then the above system of equations can be solved for X,...,X r in terms of Y,...,Y r. Consequently by Theorem 3.2. Y,...,Y r = X,...,X r = S. It follows from problem that Y,...,Y r is a basis for S. (ii) We show that all bases for S are given by equations 2. So suppose that Y,...,Y r forms a basis for S. Then because X,...,X r form a basis for S, we can express Y,...,Y r in terms of X,...,X r as in 2, for some matrix A = [a ij ]. We show A is non singular by demonstrating that the linear independence of Y,...,Y r implies that the rows of A are linearly independent. So assume x [a,...,a r ] + + x r [a r,...,a rr ] = [,...,]. Then on equating components, we have Hence a x + + a r x r = a r x + + a rr x r =. x Y + + x r Y r = x (a X + + a r X r ) + + x r (a r X + + a rr X r ). = (a x + + a r x r )X + + (a r x + + a rr x r )X r = X + + X r =. Then the linear independence of Y,...,Y r implies x =,...,x r =. (We mention that the last argument is reversible and provides an alternative proof of part (i).) 4