MATH1025 W

PRACTICE  TEST WITH SOLUTIONS
On Chapters 4, 5, and 6

1. (4 marks) Find the standard matrices for the following linear transformations:

(a) T1: R3 -> R3

w1 = 3x1 - 4x2 + x3

w2 = -x1 + 5x2 + 2x3 [T1] =

w3 = 2x1 + 3x3

(b) T2: R3 -> R2

w1 = 4x1 - 2x2 + 3x3 [T2] =

w1 = 2x1 - 5x2

2. (4 marks) Determine whether the linear transformation in question 1(a) is one-to-one.

det[T1] = 3x15 + 4x(-3 - 7) +1x(-10) = -5

Since this determinant is non-zero the transformation is one-to-one.

3. (5 marks)

(a) Find the composition T2 o T1 of the two linear transformations in questions 1(a) and (b).

[T2oT1] = [T2][T1] =

It is not commutative since the matrix product [T1][T2] is not defined.

4. (2 marks) Write down the standard matrices for the following linear transformations:

(a) A reflection about the line x = y in R2;

(b) A projection onto the yz-plane in R3.

5. (7 marks) Find the standard matrix for the linear operator  T : R3 -> R3 which reflects a vector in the xy-plane and then projects it into the xz-plane by considering the effect of this operator on the standard basis e1, e2, e3.

The linear operators have the following effect on the standard basis:

reflection in xy-plane projection into xz-plane

e1 = (1, 0, 0) -> (1, 0, 0) -> (1, 0, 0) = T(e1)

e2 = (0, 1, 0) -> (0, 1, 0) -> (0, 0, 0) = T(e2)

e3 = (0, 0, 1) -> (0, 0, -1) -> (0, 0, -1) = T(e3)

Thus the standard matrix is [T] = [T(e1), T(e2), T(e3)] =

The following are the axioms which define a vector space V where u, v and w are vectors in V and k and m are scalars:

(1) If u and v are in V then u + v is in V.

(2) u + v = v + u

(3) u + (v + w) = (u + v) + w

(4) There is a zero vector 0 in V such that u + 0 = 0 + u for all u in V.

(5) For each u in V there is a negative -u such that u + (-u) = (-u) + u = 0

(6) For any u in V and any scalar k, ku is in V.

(7) k(u + v) = ku + kv

(8) (k + m)u = ku + mu

(9) k(mu) = (km)u

(10) 1u = u

6. (3 marks) Determine whether the set of 2 by 2 matrices with non-zero determinant is a vector space with the operation + denoting matrix multiplication and the usual definition of scalar multiplication.

It is NOT a vector space since the following axioms do not hold:

(2) matrix multiplication is not commutative;

(6) fails when k = 0 since then the determinant of kA is zero

(7) fails since k(AB) is not equal to (kA)(kB)

(8) fails since (k+m)A is not equal to (kA)(mA)

It is sufficient to show one of these cases where an axiom fails in order to prove it is not a vector space.

7. (5 marks) Is the set of vectors (x, 2x) a subspace of R2? Justify your answer.

We must show that axioms 1 and 6 hold.

Let u = (x, 2x) and v = (y, 2y).

Then u + v = (x+y, 2x+2y) = (x+y, 2(x+y)) which is of the required form.

Also ku = (kx, k(2x)) = (kx, 2(kx)) which is of the desired form.

Thus the set of vectors of the form (x, 2x) is a subspace.

8. (5 marks) Are the vectors (2, -1, 5), (-1, 0, 3) and (4, 4, 2) linearly independent? Justify you answer.

The three vectors are linearly independent if

a(2, -1, 5) + b(-1, 0, 3) + (4, 4, 2) = (0, 0, 0)

has the only the trivial solution a = b = c = 0. This is true if

has a non-zero determinant. The determinant has the value 2x(-12) + 1x(-2 - 20) + 4x(-3) = -58 so the vectors are linearly independent.

9. (2 marks) If V is any vector space and S = {v1, v2, ..., vn} is a set of vectors in V, under what conditions is S a basis for V?

It is a basis if:

i) the vectors in S are linearly independent;

ii) the vectors in S span the vector space V.

10. (7 marks) Find the coordinates of 2 + 3x - x2 relative to the basis {1 + x, x - x2, 1 - 3x2} for P2.

Let a(1+x) + b(x-x2) + c(1-3x2) = 2 + 3x - x2. Thus

a + c = 2, a + b = 3, -b - 3c = -1

This system of linear equations has solution a = 2, b = 1, c = 0, i.e. the co-ordinates are (2, 1, 0).

11. (4 marks) Prove that rank(A) = rank(AT).

rank of A = dimension of column space of A = dimension of row space of A

But column space of AT = row space of A (or row space of AT = column space of A).

Hence rank of A = rank of AT.

12. (7 marks) Use the Gram-Schmidt method to transform the vectors {(1, 2, 3), (4, -1, 4)} into an orthonormal set.

Let v1 = (1, 2, 3) and

v2 = (4, -1, 4) - [(4, -1, 4).(1, 2, 3)/ v1. v1] v1 = (4, -1, 4) -[14/14] (1, 2, 3) = (3, -3, 1)

Normalize

13. (10 marks)

(a) Show that

is an orthonormal basis for R3.

Let the vectors be v1, v2. v3. Then

Thus the set of vectors is orthonormal and hence a basis for R3.

(b) Find the coordinates of (2, -2, 5) with respect to the basis in part (a).

Let the vector be v. Then the coordinates are

Thus the coordinates of v with respect to this basis are