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Unformatted text preview: Linear Algebra  Problem Set 1 A Notation: For a matrix or linear operator X , det X denotes the determinant of X , Tr( X ) the trace of X , and rank( X ) the rank of X . We will denote by X t the transpose of X , by X the conjugate of X , and by X * the conjugate transpose of X . If F is a field and n a positive integer, we will denote by M n ( F ) the set of all n n matrices with entries in F , and denote by F n the vector space of n 1 column vectors with entries in F . We will also denote by I M n ( F ) the identity matrix and by diag( 1 , 2 , , n ) M n ( F ) the diagonal matrix with main diagonal entries 1 , 2 , , n . 1. Let V be a vector space over a field F . (a) If L is a linearly independent subset of V , show that V has a basis containing L . Conclude that every nonzero vector space has a basis. (b) Let W be a subspace of V . Prove that there exists a subspace U of V such that V = U W . Solution: (a) We will use Zorns Lemma. Let S be the set of all linearly independent subsets of V containing L , which is nonempty as L S . Note that S is partially ordered by inclusion. For any chain C S , let U = [ C C C which clearly contains L . To show that U is a linearly independent subset of V , let x 1 ,x 2 , ,x n U such that 1 x 1 + 2 x 2 + n x n = 0 for some i F . Since C is a chain, there exists C C such that x i C for all i { 1 , 2 , ,n } . Since C is linearly independent, i = 0 for all i { 1 , 2 , ,n } . Hence U is a linearly indepen dent subset of V containing L which is an upper bound for C . Therefore by Zorns Lemma, S contains a maximal element B , which is a linearly independent subset of V containing L . We claim that B is a basis for V . If span( B ) 6 = V then there exists an element x V such that B = B { x } is linearly independent, and B is an element of S . This contradicts the maximality of B , and so span( B ) = V , meaning that B is a linearly independent set that spans V . That is, B is a basis for V containing L . Lastly, if V is any nonzero vector space, choose any x V with x 6 = 0. Then { x } is a linearly independent subset of V . By the previous result, V has a basis containing { x } , finishing the proof. (b) Let B W be a basis for W . Since this is a linearly independent subset of V , there exists a basis B V of V which contains B W . Let C = B V \ B W , which is another linearly independent subset of V disjoint from B W . Let U = span( C ) which is clearly a subspace of V not containing W . Now if v V , we can write a unique finite sum v = X v B V v v where v F . And so v = X w B W w w + X v C v v which shows that V = U + W . But because B V is a disjoint union of B W and C , and this expression is unique, we have that V = U W . 2. Let V and W be vector spaces over the field F , with dim V = n and dim W = m . Let T : V W be a linear transformation. Prove that there exists baseslinear transformation....
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 Spring '10
 HALL
 Linear Algebra, Determinant

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