Linear Algebra Done Right Solutions Kyle Funk July 9, 2014
Chapter 1 1. Suppose a and b are real numbers, not both 0. Find real numbers c and d such that 1/(a + bi) = c + bi Solution: Multiply the top an bottom of the left side by the conjugate: (a − bi) 1 = c + bi (a − bi) (a + bi) so that
a − bi = c + di a2 + b2 and, equating real and imaginary parts of both sides, c=
a a2 + b2
d=
−b a2 + b2
and . 2. Show that
√ −1 + 3i 2
is a cube root of 1. Solution:
and
√ ! −1 + 3i 2
√ ! √ −1 + 3i −1 + 3i = 2 2
√ ! −1 − 3i 2
1
√ ! −1 + 3i = 1. 2
3. Prove that −(−v) = v for every v ∈ V . Proof. The additive inverse of −v is −(−v), by definition, and we have v + (−v) = 0. So v is also the additive inverse of (−v). Therefore, by the uniqueness of inverses, −(−v) = v.
4. Prove that if a ∈ R, v ∈ V, av = 0, then a = 0 or v = 0. Proof. : If a = 0, then clearly av = 0. If a 6= 0, then multiply both sides of the equation by 1 a: 1 1 (av) = 0. a a By the associative property, 1 ( · a)v = 0 a so v=0
5. For each of the following subsets of F3 , which we will call V , determine whether it is a subspace of F3 : (a) {(x1 , x2 , x3 ) ∈ F3 : x1 + 2x2 + 3x3 = 0}; (b) {(x1 , x2 , x3 ) ∈ F3 : x1 + 2x2 + 3x3 = 4}; (c) {(x1 , x2 , x3 ) ∈ F3 : x1 x2 x3 = 0}; (d) {(x1 , x2 , x3 ) ∈ F3 : x1 = 5x1 }; let u = (x1 , x2 , x3 ), v = (y1 , y2 , y3 ) ∈ V and a ∈ R (a) Closed under addition: x + y = (x1 + y1 , x2 + y2 , x3 + y3 ) x1 + y1 + 2(x2 + y2 ) + 3(x3 + y3 ) = 0 (x1 + 2x2 + 3x3 ) + (y1 + 2y2 + 3y3 ) = 0 0 + 0 = 0X Closed under scalar multiplication: ax = (ax1 , ax2 , ax3 ) 2
ax1 + 2ax2 + 3ax3 = 0 a(x1 + 2x2 + 3x3 ) = 0 a0 = 0X Therefore, V is a subspace of F3 . (b) This is not a subspace: let u = (1, 0, 1) and v = (0, 2, 0). Then u + v = (1, 2, 1) 6∈ V . Therefore, it this subset is not closed under scalar multiplication and is not a subspace of F3 . (c) This is not a subspace: let u = (1, 1, 0) and v = (0, 0, 1). Then u + v = (1, 1, 1) 6∈ V . So this subset is not closed under addition, and therefore not a subspace of F3 . (d) Closed under addition: x + y = (x1 + y1 , x2 + y2 , x3 + y3 ) x1 + y1 = 5x3 + 5y3 = 5(x3 + y3 )X Closed under scalar multiplication. ax1 = a(5x3 ) = 5(ax3 )X Therefore, V is a subspace of F3 . 6. Give an example of a non-empty subset U of R2 such that U is closed under addition and taking inverses, but is not a subspace of R2 . Solution: Z (the integers) is one example. Q is another. These are closed under addition and taking inverses, but because scalars are elements of R, scalar multiplication is not closed. 7. Give an example of a subset U of R2 such that U is closed under scalar multiplication, but is not a subspace of R2 . Solution (from number 5): U = {(x1 , x2 ) ∈ R2 : x1 x2 = 0} There are obviously many examples. 8. Prove that the intersection of any collection of subspaces of V is a subspace of V . Proof. Let {Ui } denote an arbitrary collection of subspaces, and let u, v ∈ ∩Ui . Since, by the definition of intersection, u, v ∈ Ui for every i. Then u + v ∈ ∩Ui and ∩{Ui } is closed under addition. Since u ∈ Ui for every i, au ∈ Ui for every i, and au ∈ ∩Ui . So ∩{Ui } is closed under scalar multiplication. Therefore, it is a subspace of V . 3
9. Prove that the union of two subspaces of V is a subspace if and only if one subspace is contained in the other. Proof. Let U, W be subspaces of V . If one subspace is contained in the other, then either U ⊂ W or W ⊂ U . If U ⊂ W , then U ∪ W = U , which is a subspace of V . If W ⊂ U , then W ∪ U = W , which is a subspace of W. Now we will use a proof by contradiction to show that U ∪ W is a subspace of V =⇒ U ⊂ W or W ⊂ U . Assume that U ∪ W is a subspace of V and that U 6⊂ W and W 6⊂ U . Then ∃ u ∈ U such that u 6∈ W and ∃ w ∈ W such that w 6∈ U . Since u, w ∈ U ∪ W , and U ∪ W is a subspace of V , u + w ∈ U ∪ W . This implies that u + w ∈ U or u + w ∈ W . If u+w ∈ U , then u+w +(−u) = w by associative property and existence of additive inverses. So we have w ∈ U by closure under addition. If u + w ∈ W , then u + w + (−w) = u ∈ W by the same argument. we have reached a contradiction and have shown that U ∪ W is a subspace of V ⇐⇒ U ⊂ W or W ⊂ U . 10. Suppose U is a subspace of V . What is U + U ? Solution: By definition, U + U = {u + u : u ∈ U }. U ⊂ U + U because u + u ∈ U by closure under addition. U + U ⊂ U because u + u + (−u) = u ∈ U . Therefore, U + U = U . 11. Is the operation of addition on the subspaces of V commutative? Associative? Solution: The operation of addition on subspaces is commutative and associative, by the associative and commutative properties of individual vectors in V . 12. Does the operation of addition on subspaces have an additive identity? Which subspaces have additive inverses? solution: {0} is the additive identity for addition on subspaces of V because, if U is a subspace of V , U + {0} = {0} + U = U . For the additive inverse of a subspace U , we are looking for a subspace V such that U + V = {0}. Since U, V ⊃ U + V , the only subspace which satisfies this is V = {0}. 13. Prove or give a counterexample: if U1 , U2 , W are subspaces of V, then U1 + W = U2 + W implies U1 = U2 . Solution: One should suspect this is untrue because of the lack of additive inverses found in the previous problem. For a counterexample, let U1 = {0} and W = U2 be any arbitrary subspace. Then both sides are equal to W (from problem 10), but U1 6= U2 . Another counterexample: U1 = {(x, 0) ∈ R2 : x ∈ R}, U2 = {(0, y) ∈ R2 : y ∈ R}, and W = {(x, y) ∈ R2 : x, y ∈ R}. 4
14. Suppose U is the subspace of P (F) consisting of all polynomial of the form p(z) = az 2 + bz 5 , where a, b ∈ P (F). Find a subspace W of P (F) such that U ⊕ W = P (F). Solution: Let W be the set of all polynomials such that the z 2 and z 5 coefficients are 0. 15. Prove or give a counterexample: if U1 , U2 , W are subspaces of V, then V = U1 ⊕ W and V = U2 ⊕ W implies U1 = U2 . Solution: Counterexample: U1 = {(x, 0) ∈ R2 : x ∈ R}, U2 = {(0, y) ∈ R2 : y ∈ R}, and W = {(z, −z) ∈ R2 : z ∈ R}.
Chapter 2 1. Prove that if (v1 , . . . , vn ) spans V , then so does the list (v1 − v2 , v2 − v3 , . . . , vn−1 − vn , vn ) obtained by subtracting from each vector (except the last one) the following vector. Proof. Let v ∈ V . to show that v ∈ span(v1 − v2 , v2 − v3 , · · · , vn−1 − vn , vn ), we need to find a1 , . . . , an ∈ F such that v = a1 (v1 − v2 ) + a2 (v2 − v3 ) · · · + an−1 (vn−1 − vn ) + an vn . rearranging the terms, we get v = a1 v1 + (a2 − a1 )v2 + (a3 − a2 )v3 + · · · + (an−1 − an )vn + an vn . Since v ∈ span(v1 , . . . vn ), there are b1 , . . . , bn ∈ F such that v = b1 v1 + · · · + bn vn . Letting a1 = b1 , a2 = b2 + a1 , a3 = b3 + a2 , . . . shows that our list in question spans V . 2. Prove that if (v1 , · · · , vn ) is linearly independent in V , then so is the list (v1 − v2 , v2 − v1 , · · · , vn−1 − vn , vn ) obtained by subtracting from each vector(except the last one) the following vector.
5
Proof. We must show that, with a1 , . . . , an ∈ F, 0 = a1 (v1 − v2 ) + a2 (v2 − v3 ) · · · + an−1 (vn−1 − vn ) + an vn has only the trivial solution. Rearranging terms, we get 0 = a1 v1 + (a2 − a1 )v2 + (a3 − a2 )v3 + · · · + (an−1 − an )vn + an vn . Since (v1 , . . . vn ) is linearly independent, we have a1 = 0, a2 − a1 = 0, a3 − a2 = 0 . . . , implying that a1 = a2 = · · · = an = 0. 3. Suppose (v1 , . . . , vn ) is linearly independent in V and w ∈ V . Prove that if (v1 +w, . . . , vn +w) is linearly dependent, then w ∈ span(v1 , . . . , vn ). Proof. If (v1 + w, · · · , vn + w) is linearly dependent then a1 (v1 + w) + · · · + an (vn + w) = 0 has a solution other than the trivial solution. Rearranging terms, we get a1 v1 + a2 v2 + · · · + an vn + (a1 + · · · + an )w = 0 . Since (v1 . . . , v2 ) is linearly independent, we know that a1 + · · · + an 6= 0, so can write w=
n X i=1
ai vi a1 + · · · + an
We have written w as a linear combination of (v1 , . . . , vn ), and have therefore shown that w ∈ span(v1 , . . . , vn ). 4. Suppose m is a positive integer. Is the set consisting of 0 and all polynomials with coefficients in F and with degree equal to m a subspace of P (F)? Solution: This set is not a subspace of P (F) because it is not closed under addition. Any example where the highest degree terms of each polynomial cancel each other out when added together is a counterexample. This result is tied to how, when we defined degree, we added the constraint that am 6= 0. 5. Prove that F∞ is infinite dimensional. Proof. We will use proof by contradiction. Assume F∞ is finite dimensional with dimension N . Then (e1 , e2 , . . . , eN ), the list of standard basis vectors, is a basis for F∞ . So clearly (e1 , e2 , . . . , eN ) spans F∞ , but eN +1 ∈ F∞ cannot be written as a linear combination of the vectors in the list (e1 , e2 , . . . , eN ), which contradicts the definition of a basis. Therefore, F∞ is infinite dimensional. 6. Prove that the real vector space consisting of all continuous functions on the interval [0, 1] is infinite dimensional. 6
Proof. Consider the list of vectors (1, x, x2 , · · · , xn ) (with necessary constrictions on x.) This list is linearly independent for all n, which will be proven by induction: • The first case, x0 = 1,is a single vector which is linearly independent because it is not 0. • When you add xn+1 to the list (1, x, x2 , · · · , xn ), the list is still linearly independent because xn+1 6∈ span(1, x, x2 , · · · , xn ) Since this list is linearly independent for all n, the vector space is infinite dimensional. 7. Prove that V is infinite dimensional if and only if there is a sequence v1 , v2 . . . of vectors in V such that (v1 , v2 , . . . , vn ) is linearly dependent for every n. Proof. 8. Let U be the subspace of R5 defined by U = {(x1 , x2 , x3 , x4 , x5 ) ∈ R5 : x1 = 3x2 andx3 = 7x4 }. Find a basis of U . Solution: The vectors v1 = (3, 1, 0, 0, 0) v2 = (0, 0, 7, 1, 0) v3 = (0, 0, 0, 0, 1) form a basis for the subspace. 9. Prove or disprove: there exists a basis (p0 , p1 , p2 , p3 ) of P3 (F) such that none of the polynomials p0 , p1 , p2 , p3 has degree 2. Proof. Here is an example of such a basis: p0 = 1, p1 = x, p2 = x2 + x3 , p3 = x2 + 3x3 This basis is linearly independent, and spans P3 (F), which is easy to verify. No vector in this list has degree 2. This is obviously one of many examples. 10. Suppose that V is finite dimensional, with dim V = n. Prove that there exist one dimensional subspaces U1 , . . . , Un of V such that V = U1 ⊕ · · · ⊕ Un .
7
Proof. Let (v1 , v2 , . . . , vn ) be a basis for V . Let Each Uj be defined by Uj = {avj : a ∈ F}. Then clearly U1 + · · · + Un = V. Since a basis is linearly independent, the equation a1 v1 + · · · + an vn = 0 P has only the trivial solution, which implies that the only way uj = 0 where uj ∈ Uj is when all uj ’s are equal to 0. So we have U1 + · · · + Un = V and ∩Un = 0. Thus, {U1 , . . . , Un } is a set of subspaces such that V = U1 ⊕ · · · ⊕ Un .
11. Suppose that V is finite dimensional and U is a subspace of V such that dimU = dimV . Prove that U = V . Proof. Let (v1 , . . . , vn ) be a basis for U . Since dimV = dimU , V has the same number of basis vectors as U . Since the list (v1 , . . . , vn ) is linearly independent, and has the correct number of vectors, it also form a basis for V . Therefore, U = V = span(v1 , . . . , vn ) 12. Suppose that p0 , p1 , . . . pm are polynomials in Pm (F) such that pj (2) = 0 for each j. Prove that (p0 , p1 , . . . pm ) is not linearly independent in Pm (F). Proof. If pj (2) = 0 for every j, then the equation a0 p0 + a1 p1 + · · · + am pm has the solution x = 2, regardless of choice of a’s. Therefore, (p0 , p1 , . . . pm ) is not linearly independent in Pm (F) 13. Suppose U and W are subspaces of R8 such that dimU = 3, dimW = 5, and U + W = R8 . Prove that U ∩ W = {0}. Proof. We have the equation dim(U + V ) = dimU + dimV − dim(U ∩ V ). Since dim(U + V ) = 8, dimW = 5, and dimU = 3, we have 8 = 5 + 3 − dim(U ∩ V ) so that dim(U ∩ V ) = 0. This implies that (U ∩ V ) must be {0}. 14. Suppose that U and W are both five-dimensional subspaces of R9 . Prove that U ∩ W 6= {0}. 8
15. You might guess that, by analogy with the formula for the number of elements in the union of three subsets of a finite set, that if U1 , U2 , U3 are subspaces of a finite dimensional vector space, then dim(U1 + U2 + U3 ) = dimU1 + dim U2 + dim U3 − dim (U1 ∩ U2 ) − dim (U1 ∩ U3 ) − dim (U2 ∩ U3 ) + dim (U1 ∩ U2 ∩ U3 ) Prove this or give a counterexample. 16. Prove that if V is finite dimensional and U1 , . . . , Um are subspaces of V , then dim(U1 , + · · · + Um ) ≤ dimU1 + · · · + dimUm 17. Suppose V is finite dimensional. Prove that if U1 , . . . , Um are subspaces of V such that V = U1 ⊕ · · · ⊕ Um , then dimV = dimU1 + · · · + dimUm .
9
