Chapter 20. Vector Spaces and Bases. Solutions to Exercises. Exercise 20.1 Find
a basis of the hyperplane in R4 with equation x + 2y + 3z + 4w = 0. Solution: ...
Chapter 20. Vector Spaces and Bases Solutions to Exercises
Exercise 20.1 Find a basis of the hyperplane in R4 with equation x + 2y + 3z + 4w = 0. Solution: There are many answers. A valid answer consists of three vectors in the hyperplane which span the hyperplane and are linearly independent. You must check spanning and linear independence. For example, the vectors v1 = (2, −1, 0, 0),
v2 = (0, 3, −2, 0),
v3 = (0, 0, 4, −3)
all satisfy the equation of the hyperplane. To check spanning, let v = (x, y, z, w) be an arbitrary vector in the hyperplane. We want to find scalars c1 , c2 , c3 such that v = c1 v 1 + c2 v 2 + c3 v 3 . This translates into the equations 2c1 = x,
3c2 − c1 = y,
4c3 − 2c2 = z,
−3c3 = w,
which has solutions c1 = x/2, c2 = (y + x/2)/3, c3 = −w/3, so spanning is proved. For linear independence we set c1 v1 + c2 v2 + c3 v3 = 0. This translates into the equations 2c1 = 0,
3c2 − c1 = 0,
4c3 − 2c2 = 0,
−3c3 = 0,
which imply c1 = c2 = c3 = 0, proving linear independence. Exercise 20.2 Let V be the vector space of solutions of the differential equation f 00 − 2f 0 + f = 0. (a) Show that the functions ex and xex belong to V. (b) Show that the functions ex and xex are linearly independent.1 Solution: a) Differentiate the functions to show they satisfy the equation. b) Suppose c1 ex + c2 xex = 0. Then c1 + c2 x = 0. This implies c1 = c2 = 0, so the functions are linearly independent. Exercise 20.3 Let Pn be the vector space of polynomials of degree at most n. Let p0 , p1 , . . . , pn be polynomials with deg(pi ) = i. Show that {p0 , p1 , . . . , pn } is a basis of Pn . 1
In fact {ex , xex } is a basis of V, but that is harder to prove.
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Solution: Proceed as with Legendre polynomials. Let V be the span of {p0 , . . . , pn }. Each pk = ak xk + (lower powers), where ak 6= 0. Since p0 = a0 is a nonzero constant, we have 1 ∈ V. Then a1 x = p1 − (multiple of 1) ∈ V ⇒ x ∈ V, so a2 x2 = p2 − (lin. comb. of 1, x) ∈ V ⇒ x2 ∈ V, so a3 x3 = p3 − (lin. comb. of 1, x, x2 ) ∈ V ⇒ x3 ∈ V, and so on. So {1, x, x2 , . . . , xn } ⊂ V. Hence every polynomial is in V so V = Pn . This proves that {p0 , p1 , . . . , pn } spans Pn . To check linear independence, suppose c0 , . . . , cn are constants such that c0 + c1 p 1 + · · · + cn p n = 0 (i.e., is the zero polynomial). The coefficient of xn is cn an = 0 and an 6= 0, so cn = 0. Now we have c0 + c1 p1 + · · · + cn−1 pn−1 = 0. Repeating the previous argument we find cn−1 = 0, and so on, so all ci = 0 and the polynomials are linearly independent. Exercise 20.4 On Pn , we have the an analogue of the dot product, given by Z 1 hp, qi = p(x)q(x) dx. −1
We say p and q are orthogonal if hp, qi = 0. In this problem you may use without proof the fact that distinct Legendre polynomials Pk and P` are orthogonal: hPk , P` i = 0 if k 6= `. (a) Suppose f ∈ Pn is orthogonal to each Legendre polynomial Pk , for all 0 ≤ k ≤ n. Show that f = 0. (b) We know that any f ∈ Pn may be uniquely expressed as a linear combination of the Pk . Show that this unique linear combination is given by n X hf, Pk i f (x) = Pk (x). hPk , Pk i k=0
[Hint: Let ∆ be the difference of the two sides and show that h∆, Pk i = 0 for all 0 ≤ k ≤ n. Then invoke part (a). ] Solution: a) We know that {P0 , P1 , . . . , Pn } is a basis of Pn . So we can write f = c0 P0 + c1 P1 + · · · + cn P n for some scalars c0 , . . . , cn . We are given that hf, P` i = 0 for ` = 0, 1, . . . , n. So 0 = hf, P` i = hc0 P0 + c1 P1 + · · · + cn Pn , P` i = c0 hP0 , P` i + c1 hP1 , P` i + · · · + c` hP` , P` i + · · · + cn hPn , P` i = c` hP` , P` i. 2
Since hP` , P` i 6= 0 (it is the integral of a positive function) we must have c` = 0 for all `. Therefore f = 0. b) By the orthogonality of the Legendre polynomials we have for every `: n X hf, Pk i hf, P` i h∆, P` i = hf − Pk , P` i = hf − P` , P` i = 0. hP hP k , Pk i ` , P` i k=0
By part a) we have ∆ = 0, which is the desired equation. Exercise 20.5 Let v1 , . . . vk be nonzero vectors in Rn which are orthogonal with respect to the dot product. Prove that {v1 , . . . , vk } is linearly independent. Solution: Suppose some linear combination c1 v1 + · · · + ck vk = 0. For each i = 1, 2, . . . , k, we take the dot product with vi on both sides. By orthogonality we get ci hvi , vi i = 0. Since vi is nonzero we have ci = 0 for all i. Therefore {v1 , . . . , vk } is linearly independent.
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