Problem set #4. Due February 19, x 1 x 2 + x 3 + x 4 x 5 = 0 x 1 + x 3 + 2x 4 = 1 x 1 x 2 x 4 x 5 = 1.

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1 Problem set #4 Due February 19, 218 The letter V always denotes a vector space. Exercise 1. Find all solutions to 2x 1 x 2 + x 3 + x 4 x 5 = x 1 + x 3 + 2x 4 = 1 x 1 x 2 x 4 x 5 = 1. Solution. First we write down the augmented matrix associated to this system and then we row reduce it. The result is Now convert this back into a system of linear equations x 1 + x 3 + 2x 4 = 1 x 2 + x 3 + 3x 4 + x 5 = 2 and solve for the bound variables in terms of the free variables x 1 = 1 x 3 2x 4 x 2 = 2 x 3 3x 4 x 5. The general solution is therefore x 1 1 x 3 2x x 2 x 3 x 4 = 2 x 3 3x 4 x 5 x 3 x 4 = 2 + x x x 1 5 x 5 x Exercise 2. Do the vectors v 1 =, v 2 = 2, v 3 = span R 3? Are they linearly 1 1 independent?

2 Solution. There are lots of ways to go about solving this. Here is one: The calculations 1 = = = show that the standard basis vectors e 1, e 2, e 3 are in Span{v 1, v 2, v 3 }, and hence R 3 = Span{e 1, e 2, e 3 } Span{v 1, v 2, v 3 }. Because we have three spanning vectors v 1, v 2, v 3 R 3 in a space of dimension 3, they must also be linearly independent. Exercise 3. Suppose v 1, v 2, v 3, v 4 V, and set w 1 = v 1 v 2, w 2 = v 2 v 3, w 3 = v 3 v 4, w 4 = v 4. (a) Show that Span{v 1, v 2, v 3, v 4 } = Span{w 1, w 2, w 3, w 4 }. (b) Show that v 1, v 2, v 3, v 4 are linearly independent if and only if w 1, w 2, w 3, w 4 are linearly independent. Solution. Both claims follow from the relations v 1 = w 1 + w 2 + w 3 + w 4 v 2 = w 2 + w 3 + w 4 v 3 = w 3 + w 4 v 4 = w 4. (a) The above relations show that v 1, v 2, v 3, v 4 Span{w 1, w 2, w 3, w 4 }, and so Span{v 1, v 2, v 3, v 4 } Span{w 1, w 2, w 3, w 4 }. On the other hand we clearly have w 1, w 2, w 3, w 4 Span{v 1, v 2, v 3, v 4 }, and so Span{w 1, w 2, w 3, w 4 } Span{v 1, v 2, v 3, v 4 }. Thus the two spans are equal. (b) First assume that v 1, v 2, v 3, v 4 are linearly independent. Any linear relation can be rewritten as = a 1 w 1 + a 2 w 2 + a 3 w 3 + a 4 w 4 = a 1 (v 1 v 2 ) + a 2 (v 2 v 3 ) + a 3 (v 3 v 4 ) + a 4 v 4 = a 1 v 1 + (a 2 a 1 )v 2 + (a 3 a 2 )v 3 + (a 4 a 3 )v 4. The linear independence of v 1, v 2, v 3, v 4 now implies a 1 =, a 2 a 1 =, a 3 a 2 =, a 4 a 3 = from which we deduce a 1 = a 2 = a 3 = a 4 =. Thus w 1, w 2, w 3, w 4 are linearly independent.

3 Now assume that w 1, w 2, w 3, w 4 are linearly independent. Any linear relation can be rewritten as = a 1 v 1 + a 2 v 2 + a 3 v 3 + a 4 v 4 = a 1 (w 1 + w 2 + w 3 + w 4 ) + a 2 (w 2 + w 3 + w 4 ) + a 3 (w 3 + w 4 ) + a 4 w 4 = a 1 w 1 + (a 1 + a 2 )w 2 + (a 1 + a 2 + a 3 )w 3 + (a 1 + a 2 + a 3 + a 4 )w 4. The linear independence of w 1, w 2, w 3, w 4 now implies that a 1 =, a 1 + a 2 =, a 1 + a 2 + a 3 =, a 1 + a 2 + a 3 + a 4 = from which we deduce that a 1 = a 2 = a 3 = a 4 =. Thus v 1, v 2, v 3, v 4 are linearly independent. Exercise 4. Suppose v 1,..., v n V. (a) If U V is any subspace, show that v 1,..., v n U Span{v 1,..., v n } U. Use this to prove that Span{v 1,..., v n } is equal to the intersection of all subspaces containing v 1,..., v n. (b) Show that w V is a linear combination of v 1,..., v n if and only if Span{v 1,..., v n, w} = Span{v 1,..., v n }. Solution. (a) The first claim is easy: if v 1,..., v n U then, as U is closed under addition and scalar multiplication, every linear combination of v 1,..., v n U. Thus Span{v 1,..., v n } U. Conversely, if Span{v 1,..., v n } U then, as v 1,..., v n Span{v 1,..., v n }, we must have v 1,..., v n U. To show that Span{v 1,..., v n } is equal to the intersection of all subspaces containing v 1,..., v n we need to prove that v Span{v 1,..., v n } v is contained in every subspace U such that v 1,..., v n U. First suppose v Span{v 1,..., v n }. If U is any subspace with v 1,..., v n U, then also Span{v 1,..., v n } U, and hence v Span{v 1,..., v n } U. Conversely, suppose v is contained in every subspace U with v 1,..., v n U. In particular, this applies with U = Span{v 1,..., v n }, and hence v Span{v 1,..., v n }. (b) First suppose that w is a linear combination of v 1,..., v n. Then w Span{v 1,..., v n } = v 1,..., v n, w Span{v 1,..., v n } = Span{v 1,..., v n, w} Span{v 1,..., v n }. The inclusion Span{v 1,..., v n } Span{v 1,..., v n, w} is obvious, and so Span{v 1,..., v n, w} = Span{v 1,..., v n }. Now suppose Span{v 1,..., v n, w} = Span{v 1,..., v n }. Then w Span{v 1,..., v n, w} = Span{v 1,..., v n }, and so w is a linear combination of v 1,..., v n.

4 Exercise 5. For each of the following, find a proof or a counterexample: (a) Adding a vector to a linearly independent list results in a linearly independent list. (b) Deleting a vector from a linearly independent list results in a linearly independent list. (c) Adding a vector to a linearly dependent list results in a linearly dependent list. (d) Deleting a vector from a linearly dependent list results in a linearly dependent list. [ ] [ ] [ ] [ ] [ ] Solution. (a) This is false: the list,, is linearly independent, but,, is not. (b) This is true: suppose we have a linearly independent list v 1,..., v n and delete the vector v i from it. The result is still linearly independent, as any linear relation determines a linear relation c 1 v c i 1 v i 1 + c i+1 v i c n v n = c 1 v c i 1 v i 1 + v n + c i+1 v i c n v n = among the original vectors v 1,..., v n. As v 1,..., v n are linearly independent, all of these c 1,..., c i 1, c i+1,..., c n are zero. (c) This is true: if v 1,..., v n are linearly dependent then we may write c 1 v c n v n = for some scalars c i not all. If we add another vector w to the list, the resulting list is still linearly dependent, as c 1 v c n v n + w =. [ ] [ ] [ ] 1 1 (d) This is false: the list,, is linearly dependent, but becomes linearly independent if you delete the last 1 1 vector. Exercise 6. Let P 3 be the vector space of all polynomials with real coefficients and degree 3. Do the polynomials span P 3? f 1 (x) = x 3 + 2x, f 2 (x) = x 2 + x + 1, f 3 (x) = x 3 + 5, f 4 (x) = x 2 + 3x 4 Solution. These polynomials do not span P 3. For example, I claim that x 3 Span{f 1, f 2, f 3, f 4 }. Indeed, if c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) + c 4 f 4 (x) = x 3 then comparing coefficients gives c 1 + c 3 = 1 c 2 + c 4 = 2c 1 + c 2 + 3c 4 = c 2 + 5c 3 4c 4 =. Using row reduction

5 we see that this is equivalent to the system c 1 + c 4 = c 2 + c 4 = c 3 c 4 = = 1 which is obviously inconsistent.