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Question 5 For each given vector b and matrix A, determine if b e im(A) 1 -2 3 (a) b 0 A 21 3 0 5 15 (b) b A2-24 9 Question 6 Find the rank and nullity of the given linear transformations T and determine which are one-to-one and which are onto. r+ y ri+r2 Question 7 Find nullity(T) if (a) T:R R2, rank(T) 1 (b) T:RR, rank(T) 0 (c) T : Rs ? R2, rank(T)-1 Question 8 Let T : V ? V be a linear transformation and assume din(V) n. If ker(T) = 0) which of the following statements are true and which are false. (a) rank(T)-n-1 (b) inn(T) = V (c) nullity(T)=0 (d) T is onto (e) T(u) = T(v) if and only if u v for all u, v E V
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Answer #1

5. (a). Let M = [A|b] =

1

-2

3

11

2

-1

3

10

4

1

-1

4

The RREF of M is

1

0

0

2

0

1

0

-3

0

0

1

1

It implies that b = (11,10,4)T = 2(1,2,4)T-3(-2,-1,1)T+(3,3,-1)T. Hence b ? im(A) = col(A).

(b). Let M = [A|b] =

0

5

15

9

2

-2

-4

-2

3

-3

-6

-4

The RREF of M is

1

0

1

0

0

1

3

0

0

0

0

1

It implies that b = (9,-2,-4)T ? im(A) = col(A).

6. (a). T(e1) = T(1,0)T = (2,0)T and T(e2) = T(0,1)T = (0,3)T. Hence the standard matrix of T is A =

2

0

0

3

The RREF of A is I2 which implies that the columns of A are linearly independent. Hence T is one-to-one.

(b). T(e1) = T(1,0,0)T =(1,0)T,T(e2) = T(0,1,0)T =(1,0)T and T(e3) = T(0,0,1)T =(0,1)T. Hence the standard matrix of T is A =

1

1

0

0

0

1

The columns of A are not linearly independent. Hence T is not one-to-one.

(c ). T(e1) = T(1,0)T =(1,2,0)T, T(e2) = T(0,1)T =(1,1,1)T. Hence the standard matrix of T is A =

1

1

2

1

0

1

The RREF of A is

1

0

0

1

0

0

It implies that the columns of A are linearly independent. Hence T is one-to-one.

(d). T(e1) = T(1,0,0,0)T =(1,0,0)T, T(e2) = T(0,1,0,0)T =(1,1,0)T T(e3) = T(0,0,1,0)T =(0,1,1)T T(e4) = T(0,0,0,1)T =(0,0,1)T. Hence the standard matrix of T is A =

1

1

0

0

0

1

1

0

0

0

1

1

A has 4 column vectors. Since dim(R3)= 3, hence the columns of are not linearly independent. Hence T is not one-to-one.

Please post the remaining questions again.

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