Adv Alegbra HW Solutions 127

Adv Alegbra HW Solutions 127 - K of Mat n k consisting of...

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127 Solution. If the origin ( 0 , 0 , 0 ) lies on a plane H , then u = 0 and H ={ v = ( x , y , z ) R 3 : (v, n ) = 0 } , where n = (α,β,γ) is a (nonzero) normal vector; that is, H is the set of all vectors orthogonal to n . In the words of Example 4.5, H = u , and so H is a subspace of R 3 . Here are three vectors in H : p = (β, α, 0 ) ; q = ( 0 ,γ, β) ; r = (γ, 0 , α). Since n = (α,β,γ) ±= ( 0 , 0 , 0 ) , at most one of these vectors is 0. If p ±= 0 and q ±= 0, for example, it is easy to see that p , q is linearly independent. Thus, h p , q i⊂ H , so that 2 = dim ( h p , q i ) dim ( H ) . Since H ±= R 3 ,wehaved im ( H )< 3, and so dim ( H ) = 2. By Proposition 4.22, h p , q i= H . (The converse of this exercise is true, but we leave its proof for a linear algebra course.) 4.11 (i) Prove that dim ( Mat m × n ( k )) = mn . Solution. A basis consists of the matrices E ij which have entry 1 in position ij and all other entries 0. (ii) Determine dim ( S ) , where S is the subspace of Mat n ( k ) consisting of all the symmetric matrices. Solution. n + 1 2 ( n 2 n ) . 4.12 (i)
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Unformatted text preview: K of Mat n ( k ) , consisting of all the skew symmetric matrices, is a subspace of Mat n ( k ) . Solution. Now 0 is skew-symmetric, for 0 T = = − 0. If A and B are skew-symmetric, then ( A + B ) T = A T + B T = − A − B = − ( A + B ), and if α is a scalar, then (α A ) T = α( A T ) = − α A . Therefore, K is a subspace of Mat n ( k ) . (ii) Determine dim ( K ) . Solution. If A is skew symmetric, then all its diagonal entries are 0. The answer is 1 2 ( n 2 − n ) . 4.13 If p is a prime with p ≡ 1 mod 4, prove that there is a nonzero vector v ∈ F 2 p with (v, v) = 0, where (v, v) is the usual inner product of v with itself [see Example 4.4(i)]. Solution. Use the Two-squares theorem....
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