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Section 1.3, exercise 12 Prove that the upper triangular matrices form a subspace of M m × n (F) . Proof. Let W be the set of upper triangular matrices in the vector space M m × n (F) . Since M m × n (F) is a vector space, it contains a zero vector and this vector is the m × n zero matrix Z . Since Z ij = 0 for 1 i m and 1 j n , then clearly Z ij = 0 for i > j so by deﬁnition Z is upper triangular and therefore is an element of W . Next let A,B W so that A ij = B ij = 0 for i > j . Then for i > j , ( A + B ) ij = A ij + B ij = 0 + 0 = 0 so that A + B is upper triangular and therefore an element of W . Finally let c F and let A be the matrix given above. Then for i > j , ( cA ) ij = c ( A ij ) = c (0) = 0 so that cA is upper triangular and therefore an element of W. Since W contains the zero vector, is closed under vector addition, and is closed under scalar multiplication, by Theorem 1.3, W is a subspace of M m × n (F) . 1

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Section 1.5, exercise 19 Prove that if { A 1 ,A 2 ,...,A j } is a linearly independent subset of M n × n (F), then { A t 1 t 2 t k } is also linearly independent. Before proving this claim, we prove the following lemma. Lemma. Let A i M n × n (F) and c i F for 1 i j then ( c 1 A 1 + c 2 A 2 + ··· + c j A j ) t = c 1 A t 1 + c 2 A t 2 + + c j A t j . Proof. This result will be established using induction on j. For j = 1, the deﬁnition of the transpose of gives (( c 1 A 1 ) t ) ij = ( c 1 A 1 ) ji = c 1 ( A 1 ) ji = c 1 ( A t 1 ) ij for 1 i,j n so that ( c 1 A 1 ) t = c 1 A t 1 . Next, suppose the result holds for j k and consider the case when j = k + 1. For this case ( c 1 A 1 + c 2 A 2 + + c k A k + c k +1 A k +1 ) t = ([ c 1 A 1 + c 2 A 2 + + c k A k ] + c k +1 A k +1 ) t = [ c 1 A 1 + c 2 A 2 + + c k A k ] t + ( c k +1 A k +1 ) t = c 1 A t 1 + c 2 A t 2 + + c k A t k + c k +1 A t k +1 where the induction hypothesis was used after regrouping. The result thus holds for j = k + 1 and therefore for all j . We now prove the main result. Proof. Let { A 1 2 j } be a linearly independent subset of M n × n (F) and consider the equation c 1 A t 1 + c 2 A t 2 + + c j A t j = 0 2
where c i F for 1 i j. Taking the transpose of both sides of the equation gives ( c 1 A t 1 + c 2 A t 2 + ··· + c j A t j ) t = 0 t ( c 1 A t 1 ) t + ( c 2 A t 2 ) t + + ( c j A t j ) t = 0 c 1 ( A t 1 ) t + c 2 ( A t 2 ) t + + c j ( A t j ) t = 0 c 1 A 1 + c 2 A 2 + + c j A j = 0 .

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