MATH1025 AF

FW00F

Anton and Rorres

Section 5.4

1.(a) Since R2 has dimension 2 a basis will only have 2 vectors.

(b) Since R3 has dimension 3 a basis must have 3 vectors.

(c) Since P2 has dimension 3 a basis must have 3 vectors.

(d) Since M22 has dimension 4 a basis will only have 4 vectors.

3. Three vestors in R3 will be a basis if they are linearly independent. Thus we must solve the system k1u1 + k2u2 + k3u3 = 0.

(a) The coefficient matrix for this system is

which has determinant 6 and hence has only the trivial solution. Thus the vectors are linearly independent and hence a basis.

(b) The coefficient matrix for this system is

which has determinant 26 and hence has only the trivial solution. Thus the vectors are linearly independent and hence a basis.

(c) The coefficient matrix for this system is

which has determinant 0 and hence has a non-trivial solution. Thus the vectors are linearly dependent and hence not a basis.

(d) The coefficient matrix for this system is

which has determinant 0 and hence has a non-trivial solution. Thus the vectors are linearly dependent and hence not a basis.

7. We must find constants p and q such that pu1 + qu2 = w. Then (p, q) are the co-ordinates with respect to the given basis.

(a) This is the standard basis and the co-ordinates are simply (3, -7).

(b) We must solve the system of equations

2p + 3q = 1

-4p + 8q = 1

This has solution p = 5/28, q = 3/14.

(c) We must solve the system of equations

p = a

p + 2q = b

This has solution p = a, q = (b - a)/2.

9. We must find constants r, s and t such that rp1 + sp2 + tp3 = p. Then (r, s, t) are the coordinates.

(a) In this case we have the standard basis so the coordinates are (4, -3, 1).

(b) We must solve the system

r + s = 2

r + t = -1

s + t = 1

which has solution r = 0, s = 2, t = -1.

15. The row echelon form for this system is

x1 + 5x3 = 0

x2 - 7x3 = 0

x3 = 0

which has only the trivial solution so the dimension of the solution space is zero and a basis consists only of the zero vector (0, 0, 0).

Section 5.5

4. See solution in text.

5.(b) The row echelon form for this system is

x1 + x2 + 2x3 = 5

x2 + x3 = 7

Thus the solution can be written as x3 = s, x2 = 7 - s, x1 = -2 - s or in vector form

x = (-2, 7, 0) + s(-1, -1, 1)

From this we can extract the solution of the homogeneous system as x = s(-1, -1, 1).

(c) The row echelon form of this system is x1 - 2x2 + x3 + 2x4 = -1 which has solution

x2 = r, x3 = s, x4 = t and x1 = -1 + 2r - s - 2t or in vector form

x = (-1, 0, 0, 0) + r(2, 1, 0, 0) + s(-1, 0, 1, 0) + t(-2, 0, 0, 1).

From this we can extract the solution of the homogeneous system as

x = r(2, 1, 0, 0) + s(-1, 0, 1, 0) + t(-2, 0, 0, 1).

6.(c) The reduced row echelon form for this matrix is so that the solution of the homogeneous system Ax = 0 is x3 = s, x4 = t, x1 = -s+(2/7)t, x2 =-s-(4/7)t

Thus a basis for the null space is (-1, -1, 1, 0) and (2, -4, 0, 7). Note that in the last case we have multiplied the vector in the solution by 7 to get the vector with integer components.

Section 5.6

4. The rank of a matrix = dimension of the row space = dimension of the column space.

The rank of a matrix plus the nullity (the dimension of the null space) = the number of columns n.

The rank of AT = rank of A (the row space and column space are interchanged on transposition). Thus if A is an m by n matrix, AT is an n by m matrix so the rank plus the nullity of AT = m

Thus we have

(a) rank = 3 = dimension of row space = dimension of column space

Since m = n = 3 the nullity of A and AT are both 0.

(b) rank = 2 = dimension of row space = dimension of column space

Since m = n = 3 the nullity of A and AT are both 1.

(c) rank = 1 = dimension of row space = dimension of column space

Since m = n = 3 the nullity of A and AT are both 2.

(d) rank = 2 = dimension of row space = dimension of column space

Since m = 5 and n = 9 the nullity of A is 7 and the nullity of AT is 3.

(e) rank = 2 = dimension of row space = dimension of column space

Since m = 9 and n = 5 the nullity of A is 3 and the nullity of AT is 7.

(f) rank = 0 = dimension of row space = dimension of column space

Since m = n = 4 the nullity of A and AT are both 4.

(g) rank = 2 = dimension of row space = dimension of column space

Since m = 6 and n = 2 the nullity of A is 0 and the nullity of AT is 4.

5. The largest possible rank of an m by n matrix is the minimum value of m and n.

The largest possible rank gives the smallest possible nullity.

(a) Since A is 4 by 4 the largest possible rank is 4, the smallest possible nullity is 0.

(b) Since A is 3 by 5 the largest possible rank is 3, the smallest possible nullity is 2.

(c) Since A is 5 by 3 the largest possible rank is 3, the smallest possible nullity is 0.

6. From question 5, the largest possible rank of an m by n matrix is min(m, n) and the smallest possible nullity is n - min(m, n).

Section 6.3

3. (a) the second and third vectors are not orthogonal;

(b) all vector are orthogonal;

(c) the second and third vectors are not orhtogonal

(d) all vectors are othogonal

4. All the vectors in question 3 have unit length but only sets (b) and (d) are orthogonal and hence orthonormal.

9. Since the vectors v1, v2 and v3 are an orthonormal set and there are 3 of them they form an orthonormal basis for R3. Thus any vector can be expressed as v = av1 +bv2 +cv3 where a = v.v1, b = v.v2, c = v.v3.

(a) a = -7/5, b = 1/5, c = 2.

(b) a = -37/5, b = -9/5, c = 4.

(c) a = -3/7, b = -1/7, c = 5/7.

17. (a) Since u1 and u2 are orthogonal we can take v1 = u1 and v2 = u2. Thus

v3 = u3 - (u3.v1/v1.v1) v1 - (u3.v2/v2.v2) v2 = (1, 2, 1) - (4/3)(1, 1, 1) - (1/2)(-1, 1, 0) = (1/6, 1/6, -1/3).

Now normalize the vectors to get

(b) Take v1 = u1 then v2 = u2 - (u2.v1/v1.v1) v1 = (3, 7, -2) - 3(1, 0, 0) = (0, 7, -2) and

v3 = u3 - (u3.v1/v1.v1) v1 - (u3.v2/v2.v2) v2 = (0, 4, 1) - 0(1, 0, 0) - (26/53)(0, 7, -2) =

(0, 30/53, 105/53).

Now normalize the vectors to get