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CM222A LINEAR ALGEBRA Solutions 1 1. Determine whether the following sets of vectors are linearly independent subsets of the indicated vector space V. (i) (ii) (iii) {(1, −1, 4), (1, 1, 5), (1, −5, 2)}, V = R3 . {(2, 4, −4, 4), (2, −1, 0, 1), (−2, 3, −2, 2), (1, −1, −2, 4)}, V = R4 . {(3, 0, 2, −1), (4, −1, 1, −4), (2, 1, 1, 0), (3, −2, 3, −2)}, V = R4 . (i) Put vectors in rows and reduce to echelon form. R20 = R2 − R1 1 −1 4 1 −1 4 1 −1 4 0 R3 = R3 + 2R2 0 1 0 2 1 1 5 R30 = R3 − R1 2 1 −→ 0 0 0 1 −5 2 0 −4 −2 −→ and since the last row is zero, the vectors are linearly dependent. [Of course, simply writing 3(1, −1, 4) − 2(1, 1, 5) − (1, −5, 2) = (0, 0, 0) would be a correct answer, but that would hide the way the the work is done.] R20 = R2 − R1 2 4 −4 4 2 4 −4 4 2 −1 0 −5 R30 = R3 − R1 0 1 4 −3 (ii) 1 0 2 3 −2 2 0 −1 2 −2 R4 = R4 − 2 R1 1 −1 −2 4 0 −3 0 2 −→ 2 4 −4 4 2 4 −4 4 R2 ↔ R3 then 0 0 −1 0 −1 2 −2 2 −2 R30 = R3 − 5R2 R4 = R4 − R3 0 0 0 −→ R4 = R4 − 3R2 0 −6 7 0 −6 7 −→ 0 0 −6 8 0 0 0 1 so with no zero rows, the vectors are linearly independent. (iii) The same procedure as in (ii) results in a row of zeros and so shows that these vectors are linearly dependent. 2. For what real values of a and b is {(1, b, 0), (a, a, 1), (0, 0, 1 + a)} a linearly dependent subset of R3 ? 1 b 0 1 b 0 0 R2 = R2 − aR1 a a 0 a − ab 1 . 1 −→ 0 0 1+a 0 0 1+a If a = 0 or b = 1 the operation R30 = R3 − (1 + a)R2 results in a zero row. If a = −1 the last row is already zero. Hence the vectors are linearly dependent if a = 0 or −1 or if b = 1. a x e 3. Let A = t i c and let e1 , e2 , e3 be the usual basis. Write down : h m s e3 t Ae2 , (Ae1 )t , e1 t Ae3 , e3 t Ae2 , e1 t Ae1 , e2 t A, e3 t Ae3 . The answer is : m, ath, e, m, a, tic, s. 4. (i) Find a basis of R3 containing (2, 1, 3) and (1, 1, 1). Find a basis of R5 containing (1, 4, 7, 0, 3), (2, 9, 13, 2, 6) and (1, 5, 6, 2, 8). µ ¶ µ ¶ 1 1 1 R20 = R2 − 2R1 1 1 1 (i) Note that . Clearly adding (0, 0, 1) to these vectors 2 1 3 −→ 0 −1 1 gives 3 vectors in echelon form with non-zero entries in all the diagonal positions and hence a basis. Therefore {(2, 1, 3), (1, 1, 1), (0, 0, 1)} is a basis of R3 . [There are other correct answers.] 1 4 7 0 3 R20 = R2 − 2R1 1 4 7 0 3 1 4 7 0 3 0 R3 = R3 − R2 R30 = R3 − R1 0 1 −1 2 0 0 1 −1 2 0 . (ii) 2 9 13 2 6 −→ 1 5 6 2 8 −→ 0 1 −1 2 5 0 0 0 0 5 (ii) Thus suitable vectors to add are (0, 0, 1, 0, 0) and (0, 0, 0, 1, 0). 5. Prove that a set S of vectors of a vector space V is a basis if and only if (i) S is linearly independent, and (ii) every subset of V which properly contains S is linearly dependent. Note: Properly contains means “contains and is not equal to”. A set satisfying (i) and (ii) above is said to be a maximal linearly independent set. This is little more than a paraphrase of the definition of a basis. If S = {v1 , v2 , . . . , vn } is a basis it is certainly linearly independent. Also if a set T properly contains S then there is a vector / S. If w = 0 thenPT is a linearly dependent set. Otherwise, since S is Pnw ∈ T with w ∈ n spanning, w = 1 αi vi for some scalars αi and w − 1 αi vi = 0 shows that T is linearly dependent. Conversely, if S satisfies (i) and (ii) it is linearly independent. To show that it is spanning, let v be any vector.PIf v ∈ S then clearly v ∈ span S. If v ∈ / S then S ∪ {v} is linearly dependent so for vi ∈ S we have αv + αi vi = 0 for some scalars α, αi which are not all 0. We cannot have α = 0 for then S would be linearly dependent. Therefore we can divide by α and rearrange the above relation to show that v ∈ span S. Thus S is spanning. [Alternatively one could say that, since S is linearly independent, by Theorem 1.3 it can be extended to a basis. But from (ii) any strictly larger set cannot be a basis, so S itself must be a basis.] 2006/07