SECTION 3.3. PROBLEM 22. The null space of a matrix A is: N(A) = {X : AX = 0}. Here are the calculations of AX for X = a,b,c,d, and e. =

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1 SECTION 3.3. PROBLEM. The null space of a matrix A is: N(A) {X : AX }. Here are the calculations of AX for X a,b,c,d, and e. Aa [ ][ ] 3 3 [ ][ ] Ac 3 3 [ ] 3 3 [ ] Ae [ ], Ab [ ][ ] [ ] [ ][ ], Ad 3 3 ][ ] [ ] [ 3 3 [ ] [ ] + 3+ Therefore, the vectors a,c, and e are in N(A), but the vectors b and d are not. PROBLEM 4. Here are the calculations of AX for X v,w,x,y, and z. [ ] 3 Av [ ] [ ++ ] [ ], Aw [ ] [ + ] [ ], Ax [ ] [ + ] [ ], Ay [ ] [ 4 + ] [ ] Az [ ] [ ++ ] [ ] Hence, the vectors v,w, and z are in N(A), but the vectors x and y are not. PROBLEM 34. The null space of a matrix A is: N(A) {X : AX }. We can find a simple algebraic specification for N(A) by reducing A to its reduced echelon form: 7 A 3 5, 3, E [ ] 3

2 The last matrix E is in reduced echelon form. Since N(A) N(E), we have the following algebraic specification for N(A): x +7x 3, x +3x 3 To find an algebraic specification for R(A), we consider the augmented matrix [A b], where b b b b 3 Row-reduction gives: b [A b] 3 5 b, 7 b 3 b 3 b b, 6 b 3 b b 3 (b 3 b ) (b b ) The last matrix is in echelon form. The matrix equation AX b can be solved if and only if (b 3 b ) (b b ). This simplifies to 3b b +b 3 This equation is the algebraic specification for R(A). PROBLEM 36. The null space of a matrix A is: N(A) {X : AX }. We can find a simple algebraic specification for N(A) by reducing A to its reduced echelon form: A,, 3 Thus, we see that A has rank 3 and so is a nonsingular matrix. The matrix equation AX has only one solution, namely X. Hence N(A) {} Also, since A is nonsingular, it follows that for every vector b in R 3, the matrix equation AX b has at least one solution (in fact, exactly one solution). Therefore, R(A) R 3

3 SECTION 3.4. PROBLEM. We solve this system of equations by Gauss-Jordan elimination. Sunstracting the nd equation from the st equation gives the following equivalent system of equations: x +x 3 +x 4 x x 3 x 4 Thus, the leading variables are x and x. The free variables are x 3 and x 4. The solutions to the given system of equations are given in vector form by: x x 3 x 4 X x x 3 x 3 +x 4 x 3 x 3 +x 4 x 4 x 4 where x 3 and x 4 are arbitrary numbers. In this question, W is a subspace of R 4 and has the following basis:, PROBLEM 4. We just have a system of equation in the 4 unknowns x,x,x 3 and x 4. Note that x is the leading variable and x,x 3 and x 4 are the free variables. Here is a description of the solutions in vector form: x X x x 3 x 4 x x 3 x x 3 x 4 x +x 3 +x 4 In this question, W is a subspace of R 4 and has the following basis:,,

4 PROBLEM b. The vector x is in W because x,x 3,x 3,x 4 is a solution to the homogeneous system of equations given in exercise. That is, +3 +( ) 3 () ( ) In order to express x as a linear combination of the basis vectors found in exercise, we just consider the vector equation a +b 3 3 We get the solution x by letting a x 3 and b x 4. Thus, 3 PROBLEM c. The vector 7 x 8 3 is not in W because x 7,x 8,x 3 3,x 4 is not a solution to the system of equations given in exercise. The first equation in the system is not satisfied because PROBLEM b. We can find a basis for the null space of A by finding the reduced echelon form for A. We do the row-reduction as follows: A 3 5 8,,

5 3, E The last matrix E is the reduced echelon form for A. The solutions to AX are the same as the solutions to EX. Denoting the unknowns by x,x,x 3,x 4, the solutions to EX are described by the equations x x 3 x 4, x x 3 +x 4 where x 3 and x 4 are arbitrary numbers. In vector form, the solutions are given by x x 3 x 4 X x x 3 x 3 +x 4 x 3 x 3 +x 4 x 4 x 4 In this question, W N(A) is a subspace of R 4 and has the following basis:, PROBLEM b. We can find a basis for the null space of A by finding the reduced echelon form for A. We do the row-reduction as follows: A,, E 3 5 The last matrix E is the reduced echelon form for A. The solutions to AX are the same as the solutions to EX. Denoting the unknowns by x,x,x 3, the solutions to EX are described by the equations x x 3, x x 3 where x 3 is an arbitrary number. In vector form, the solutions are given by x x 3 X x x 3 x 3 x 3 x 3 In this question, W N(A) is a subspace of R 3 and has the following basis:

6 PROBLEM 3. We first determine if the matrix A which has the three given vectors as its columns is singular or nonsingular: A 3,, / /, 3 / ThelastmatrixisinechelonformandsotherankofAis3. ThismeansthatAisnonsingular. Thus, the set S is a linearly independent set consisting of three vectors and therefore S is a basis of R 3. PROBLEM 33. As in problem 33, we first determine if the matrix A which has the three given vectors as its columns is singular or nonsingular: A 5 3, 3, 6 4 /3, 6 4 /3 The last matrix is in echelon form. The rank of A is. Hence A is a singular matrix. The set S is a linearly dependent set. Hence S is not a basis for R 3. PROBLEM 34. The matrix A which has the four vectors in S as its columns is a 3 4 matrix. Hence, rank(a) 3 and hence the rank of A cannot be four. The set S is a linearly dependent set and so S isn t a basis for R 3. PROBLEM 35. The set S consists of two vectors in R 3. Those two vectors determine a plane π in R 3 through the origin. It is clear that Sp(S) is the set of vectors which lie on that plane π. Thus, Sp(S) R 3. Therefore S is not a spanning set for R 3. Therefore S cannot be a basis for R 3. SECTION 3.5. PROBLEM. Since neither vector in the set S {u,u } is equal to scalar multiple of the other, the set S is a linearly independent set of two vectors in R. Using part 3 of theorem 9, it follows that S is a basis for R.

7 PROBLEM. Since neither vector in the set S {u,u 3 } is equal to scalar multiple of the other, the set S is a linearly independent set of two vectors in R. Hence S is a basis for R. PROBLEM. The set S contains three vectors in R 3. This set S will be a basis for R 3 if and only if the S is a linearly independent set. We test this by determining whether the matrix A which has the vectors in S as its columns is singular or nonsingular: A,,, Thus, A has rank 3 and so A is nonsingular. Therefore, S is a basis for R 3. PROBLEM 3. We are given a set S of three vectors in R 3. We will consider the matrix A which has the three vectors in S as its columns. We determine the rank of A as follows: A 3, 3, 3, 4 The last matrix is in echelon form and so A has rank. This implies that S is a linearly dependent set. Therefore, S is not a basis for R 3. PROBLEM. First, we find the reduced echelon form for A: [ ] [ ] [ ] A,, 5 5 [ ] [ ], E The matrix E is the reduced echelon form for A. The solutions to the matrix equation AX are the same as the solutions to the matrix equation EX. That is, N(A) N(E). If we denote the unknowns by x,x,x 3, then x and x are the leading variables and x 3 is a free variable. The solutions to AX are described by x x 3,x x 3. In vector form, the solutions are described by: x x 3 X x x 3 x 3 x 3 x 3

8 A basis for N(A) is: Hence dim(n(a)) and so the nullity of A is. The rank of A is because the reduced echelon form E for A has nonzero rows. PROBLEM 4. The null space of A is a subspace of R 4. To find a basis for N(A), we first find the reduced echelon form for A: 5 5 A 3 7, 3 9, E Denoting the unknowns by x,x,x 3,x 4, then x,x and x 4 are the leading variables and x 3 is the free variables. The matrix equation EX is equivalent to the system of equations: x x 3 x +x 3 x 4 Thus, the solutions to EX can be described in vector form as follows: x x 3 X x x 3 x 3 x 3 x 3 x 4 Thus, a basis for N(A) is: The nullity of A is. The rank of A is 3.

9 PROBLEM 6. To find a basis for R(A), we use the method of casting out vectors. To do this, we first find the reduced echelon form for A: A 4 4, 5 4, 4 3,, E Since the leading s in E occur in columns and, we obtain a basis for R(A) by choosing those two columns from A. Here is a basis for R(A):, 4 Thus, dim(r(a)). Since dim(r(a)) rank(a), we have rank(a). Therefore, the nullity of A is dim(n(a)) 4 - rank(a) 4 -. PROBLEM A: This is question from the Winter, 8 sample exam. The matrix A 4 3 has E for its reduced echelon form. (a) To find a basis for N(A), we use the fact that N(A) N(E). This is a subspace of R 5. Letting x,...,x 5 denote the entries of a vector X, we must describe the solutions to EX, where is the zero-vector in R 3. That matrix equation is equivalent to the system of equations x +x +x 3 +x 4 +x 5, x 4 x 5. Thus, x and x 4 are the leading variables; x,x 3, and x 5 are the free variables. We can describe the solutions in vector form as follows: x x x 3 x 5 x X x 3 x 4 x x 3 x 5 x +x 3 +x 5 x 5 x 5

10 where x,x 3 and x 5 are arbitrary. As explained in class, we can now find a basis for N(A). The set,, is a basis for N(A). (b) The reduced echelon form for A has two nonzero rows. Hence rank(a). Hence R(A) has dimension. We can find a basis for R(A) by simply choosing any linearly independent set consisting of two vectors in R(A). Each column of the matrix A must be in R(A). It is easy to choose two of those columns to form a linearly independent set. For example, we can choose the first and fourth columns. This gives us the following basis for R(A):, Alternatively, the method of casting out vectors, as described in class, tells us that we can find a basis for R(A) by picking certain columns from the matrix A. The leading s in E occur in the first and fourth columns of E. The corresponding columns of A will give us a basis for R(A). It is the same basis for R(A) that we have already found. (c) The dimension of N(A) is equal to 3 because the basis for N(A) found above consists of three vectors. The dimension of R(A) is equal to because the basis for R(A) found above consists of two vectors. (d) We can choose the following set as an example of a spanning set for R(A): S, 3, The third vector is the sum of the first two vectors listed. Thus, the set S is a linearly dependent set. The subspace spanned by the set S is actually the same as the subspace spanned by just the first two vectors listed. That subspace is precisely R(A). Hence S is a spanning set for R(A), but cannot be a basis for R(A) since S is a linearly dependent set.

11 PROBLEM B. Let A denote the 6 zero matrix. The null space for A is a subspace of R 6. By definition, N(A) {X : AX }. Notice that if X is any vector in R 6, then [ ] [ ] AX X Thus, every vector X in R 6 satisfies that equation AX. Therefore, The null space of A is R 6. N(A) R 6