Linear Algebra 1A - solutions of ex. 8

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1 Linear Algebra 1A - solutions of ex. 8 Sum of subspaces, direct sum (skhum yashar) 1) Let V be a vector space, and let U, W V be two subspaces. (a) Prove that V = U W if and only if the following two statements hold. (i) V = U + W, and (ii) The only way to represent the zero vector in V as a sum of a vector from U with a vector in W is = +. (b) Consider the subspace S = U + W of V. Prove that S is a direct sum of U and W if and only if there exists at least one vector x S that has unique representation of the form x = u + w, where u U, w W. - proofs- (a). Suppose that V = U W. Then (i) holds, because V is the sum of U and W (who cares that it is direct? (i.e., that we also have U W = {})). Let us prove (ii). Assume that we can represent the zero vector of V as = u + w, where u U, w W. Then W w = u U (because U is a subspace, so is closed under addition.) But then w U W = {}, hence w =. Similarly, u =. So the only way to represent as u + w with u U, w W is with u =, w =. (And it is indeed = + is valid.). Suppose (i), (ii) hold. First, V = U + W, so we need to prove that U W = {}. Suppose v U W ; then ( v) is also in U W. We can represent the zero vector of V as v+( v) =, so (since v U, ( v) W ), v = by (ii). Hence, U + W is a direct sum. (b). Suppose S = U W. Then by part (a), the zero vector has a unique such representation.. Let us use part (a): it is enough to prove (i) and (ii) for S, U and W. In fact, (i) is given, so we only need to prove (ii). Suppose there exists a vector x S with a unique representation of the form x = u + w (#) with u U, w W. We claim that the representation of the zero vector is also unique. Suppose u U, w W satisfy = u + w. By adding zero to both sides of (#) we get x = u+u +w +w. But then we get a representation of x, and since we know it is unique, we have u + u = u, w + w = w, hence u =, w =. So (ii) from part (a) holds. This completes the proof. 2) In F 3, let W = {(, b, c) : b, c F} and U = {(a, a, a) : a F}. Prove that F 3 = U W. (Here F is a eld) 1

2 - proof - We should prove that F 3 = U + W (i.e., that their sum gives all F 3 ) and that the sum is direct. Any vector (a, b, c ) can be written as a sum { U }} { { W }} { (a, b, c ) = (a, a, a ) + (, b a, c a ) U + W. Thus, F 3 U + W F 3, so F 3 = U + W. The sum is direct, since U W = {(a, b, c) : a = and a = b = c} = {(,, )} = {}. Remark: Another way to prove the rst part is nding that {(, 1, ), (,, 1)} is a spanning set of W (actually, a basis, but it is not needed), and {(1, 1, 1)} is a spanning set of U. Hence, U + W = span{(, 1, ), (,, 1), (1, 1, 1)}. This last can be proven to be all F 3 (for instance, span{(, 1, ), (,, 1), (1, 1, 1)} = {s(, 1, ) + t(,, 1) + q(1, 1, 1) : s, t, q F} = = {(q, s+q, t+q) : s, t, q F} = {(s+q)(, 1, )+(t+q)(,, 1)+q(1,, ) : s, t, q F} = = span{(, 1, ), (,, 1), (1,, )} = F 3.). 3) Let V = Mat nxn (F) be the vector space of all nxn matrices with entries in F, where char(f) 2. Let U V denote the subspace of all symmetric matrices, and let W V denote the subspace of all antisymmetric matrices. Prove that V = U W. (A matrix A is symmetric if A t = A, and antisymmetric if A t = A.) Hint: ex. 3, q.7(e). - proof - By ex.3, q.7(e), any matrix in Mat nxn (F), where char(f) 2, can be written as a sum of a symmetric and an anti-symmetric matrices, so we already know that Mat nxn (F) = V U + W V, hence V = U + W. Thus, it is left to prove that the sum is direct. Indeed, for a eld with char 2, the only matrix that is both symmetric and anti-symmetric is the zero matrix, since U W = {A V : A = A t, A = A t } = {A V : A t = A t } = {A V : 2A t = } }{{} = {A V : A t = } = {}. char(f) 2 4) Let U, V and W be three subspaces of a certain vector space. Prove that (U V ) + (U W ) U (V + W ) Give an example in R 2 that the inclusion (hakhala) can be proper (mamash; hakhala mamash). - solution - Suppose x (U V ) + (U W ), that is, x is of the form x = y + z with y U V, z U W. 2

3 First, y and z are both in U, so y + z U (since U is a vector space). Also, y V, z W, so y + z V + W. So nally, we have x = y + z U (V + W ). Hence, (U V ) + (U W ) U (V + W ). Linearly dependent (tluim linearit) or independent (bilti tluim linearit) sets 5) Let v 1,..., v n be vectors in a vector space V. Prove that v 1,..., v n are linearly dependent if and only if for some 1 i n the vector v i can be expressed as a linear combination (tzeruf/combinatziya linearit) of the other vectors. Is it true that if v 1,..., v n are linearly dependent, the each v i can be expressed as a linear combination of the others? (prove or disprove.) - solution -. Suppose v 1,..., v n are linearly dependent. So there exist α 1,..., α n not all zero, such that α 1 v α n v n = ; So we know that for some 1 k n,. Hence, we can rewrite the sum α 1 v v k 1 + v k + +1 v k α n v n = as v k = α1 v v 1 +1 v 1... αn v n. Thus, v k is a linear combination of the other vectors.. Suppose v i is a linear combination of the other vectors: v i = α 1 v α i 1 v i 1 + α i+1 v i α n v n. Hence, α 1 v α i 1 v i 1 1 v i + α i+1 v i α n v n =. So, there is a non-trivial (not all coecients are zero) linear combination of the vectors, which (the combination) equals zero, so the vectors are dependent. About the additional question: this claim is false! In case we have a linearly dependent system, we can only say that there exists some vector, but cannot claim that all the vectors are linear combinations of the others. For example, in R 3, in the linearly dependent system (1,, ), (, 1, ), (, 2, ), the last vector is a linear combination of the others: (, 2, ) = (1,, ) + 2 (, 1, ), but (1,, ) cannot be written as a linear combination of the other two vectors. 6) Check the following sets of vectors for linear independence: (a) (1, 1, 1, ), (1, 1,, 1), (1,, 1, 1), (4, 1, 1, 3). Let us check: write these vectors as rows of a matrix and bring this matrix to its reduced echelon form (no need to make it canonical) Thus, we get four linearly independent vectors in the rows of the reduced echelon form (why?), so the initial four vectors were also linearly independent. ; 3

4 (The initial set of vectors spans the same subspace of F 4 as the rows of the reduced echelon matrix.) (b) (1, i) and (i, 1) in C 2 (over C). These are not linearly independent over C, since (i, 1) = i (1, i). (Formally, one can perform the same precedure as in (a), and get a matrix with a zero row.) (c) t 2 + 3t 2, t + 5, t 2 t + 1 in P 3 (R)[t] (the vector space of polynomials in t with degree not more than 3 over the reals). Denote p 1 (t) = t 2 + 3t 2, p 2 (t) = t + 5, p 3 (t) = t 2 t + 1; We need to check whether these three vectors are linearly independent. Suppose α 1, α 2, α 3 R satisfy α 1 p 1 +α 2 p 2 +α 3 p 3 = ( this zero denotes the zero polynomial!). That is, α 1 (t 2 + 3t 2) + α 2 (t + 5) + α 3 (t 2 t + 1) = (#) ; Rewriting this, (α 1 + α 3 )t 2 + (3α 1 + α 2 α 3 )t + ( 2α 1 + 5α 2 + α 3 ) = ; Two polynomials are equal if and only if all coecients of corresponding powers of the variable t are equal. That is, the equality holds if and only if α 1 + α 3 = 3α 1 + α 2 α 3 = ; 2α 1 + 5α 2 + α 3 = Let us write this system of linear equations (on α 1, α 2, α 3 ) in a matrix form: This (and the initial) system is an homogeneous system and has only the trivial solution (because there are no free variables, or because it's row-equivalent to the identity matrix). Thus, the only solution of the equation (#) is α 1 = α 2 = α 3 = ; Hence, the system of vectors (the three polynomials) are linearly independent in P 3 (R)[t]. ( ) ( ) ( ) (d),, in M x2 (R). As( in (c), suppose ) ( α 1, α 2, ) α 3 R ( satisfy ) α 1 + α α 1 3 = = 7 That is, we have 2α 1 + α 2 + 7α 3 = 5α 1 + α 2 + 7α 3 = ; α 1 + α 2 + α 3 = 2α 1 + α 2 + 7α 3 = This system is equivalent to the system { 2α 1 + α 2 + 7α 3 = 5α 1 + α 2 + 7α 3 = ; ( ) (#); ; 4

5 Thus, we have a homogeneous system with 3 variables and 2 equations, hence, it has a non-trivial solution. So there exist a non-trivial solution to (#) (α 1, α 2, α 3 R not all zero), hence, the given system of vectors (matrices in this case) is linearly dependent. 7) Let V be a vector space, and K V a nite subset. (a) Prove that if K contains the zero vector, then K is linearly dependent. (b) Prove that if K contains two equal vectors, then it is also linearly dependent. (c) Prove that if K is linearly independent, then any subset S of K (S K) is also linearly independent. (d) Prove that if K is linearly dependent, then any subset of V that contains K is also linearly dependent. - Proofs - (a) Suppose K = {v 1, v 2,..., v q } and suppose (WLOG) that v 1 =. Then we have 1 v 1 + v v q =. Thus, there is a non-trivial linear combination of these vectors, that gives the zero vector, so the system is linearly dependent. (b) Suppose K = {v 1, v 2,..., v q }, and (say) v 1 = v 2. Then we can write 1 v 1 + ( 1) v 2 + v v q =. So again, the given system of vectors is linearly dependent. (c) Suppose K = {v 1, v 2,..., v q } is linearly independent, and take S T a subset of K. Suppose S = {v 1, v 2,..., v r }, where r q. (We can take such S, as otherwise we would rename the vectors of K in such a way, that the vectors of S appear in the beginning. This would not eect linear dependency or independency of anything, as these does not depend on the order of vectors.) So, suppose that S is linearly dependent: α 1, α 2,..., α r F, not all equal zero, such that α 1 v 1 + α 2 v α r v r =. But then we can write α 1 v 1 + α 2 v α r v r + v r v q = ; In this last equality, not all coecients of the vectors are zero, hence we get that K is linearly dependent, while we were given that it is linearly independent. Contradiction! So S should also be linearly independent. (d) Suppose that K is linearly dependent, and that T V is a nite subset contains K (this should have been stated in the question, as we only talk about nite systems of linearly (in)dependent vectors). By the previous part ((c)), if T is linearly independent, the set K, which is a subset of T should also be linearly independent, which contradicts our assumption. So T must be also linearly deoendent. 8) Let S be a set of m vectors in F n : v 1 = (v 11, v 12,..., v 1n ), v 2 = (v 21, v 22,..., v 2n ),..., v m = (v m1, v m2,..., v mn ). Let q be a positive integer smaller than n. Pick any q dierent coordinate numbers between 1 and n 5

6 (that is, choose some ordered set 1 k 1 < k 2 <... < k q n). In each vector v 1,..., v m, pick only the coordinates in the entries of these q chosen numbers (the same q coordinates of each vector). You'll get a set S of m vectors, this time in F q : w 1 = (v 1k1, v 1k2,..., v 1kq ), w 2 = (v 2k1, v 2k2,..., v 2kq ),..., w q = (v qk1, v qk2,..., v qkq ) F q. (! - see a correction.) Prove that if the set S is linearly dependent, then this new set of shortened vectors, S, is also linearly dependent. (!) Correction: the last line should contain m vectors, as it is declared in the previous line (and as the whole question is written) : w 1 = (v 1k1, v 1k2,..., v 1kq ), w 2 = (v 2k1, v 2k2,..., v 2kq ),..., w m = (v mk1, v mk2,..., v mkq ) F q. (As it is described above, we take m vectors in F n and take only q coordinates of each (same coordinates), so from each vector we get a shortened vector, thus m vectors in F q.) - Proof - Suppose that v 1,..., v m are linearly dependent. That is, α 1, α 2,..., α m F such that α 1 v 1 + α 2 v α m v m =. This equality should hold in each one of the n coordinates, so we have α 1 v 11 + α 2 v α m v m1 = α 1 v 12 + α 2 v α m v m2 = α 1 v 1n + α 2 v 2n α m v mn = Well, let us consider only the q coordinates that we chose: 1 k 1 < k 2 <... < k q n : we have α 1 v 1k1 + α 2 v 2k α m v mk1 = α 1 v 1k2 + α 2 v 2k α m v mk2 = α 1 v 1kq + α 2 v 2kq α m v mkq = So, we get that for each one of the q coordinates of the set of vectors w 1,..., w m there is a linear combination (of the corresponding coordinates) (with the same alphas everywhere) that equals zero. If it holds for each coordinate, with the same coecients, we have that actually α 1 w 1 + α 2 w α m w m =. And, not all alphas are zero, so the set S = {w 1,..., w m } of shortened vectors is also linearly dependent. 6