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**Unformatted text preview: **Math 294 - HW5 Solutions 4.1 Introduction to Linear Spaces 4. The subset V is a subspace of P 2 : • The neutral element f ( t ) = 0 (for all t ) is in V since integraltext 1 dt = 0. • Let f and g be in V . By the definition of V , integraltext 1 fdt = 0 and integraltext 1 gdt = 0. We have to check that f + g is in V (i.e. integraltext 1 ( f + g ) fdt = 0). Consider integraltext 1 ( f + g ) fdt = integraltext 1 fdt + integraltext 1 gdt = 0, and f + g is in V . • Let f be in V and k be any constant. We see that integraltext 1 kfdt = k integraltext 1 fdt = 0, and kf is in V . Let f ( t ) = a + bt + ct 2 be in V . Since f is in V , by definition, integraltext 1 fdt = 0 ⇒ integraltext 1 fdt = bracketleftbig at + b 2 t 2 + c 3 t 3 bracketrightbig 1 = a + b 2 + c 3 = 0 ⇒ a = − b 2 − c 3 . So, elements of V are polynomials whose coefficients satisfy a = − b 2 − c 3 . That is, f ( t ) = ( − b 2 − c 3 ) + bt + ct 2 = b ( t − 1 2 ) + c ( t 2 − 1 3 ) , and f ( t ) can be described as a linear combination of t − 1 2 and t 2 − 1 3 . Since f ( t ) is an arbitrary element of V , t − 1 2 , t 2 − 1 3 form a basis of V . 8. The set V of upper triangular 3 × 3 is a subspace of R 3 × 3 . Recall a matrix is upper triangular if there are only zeros below the main diagonal (i.e. a ij = 0 for i > j ). • The zero matrix is in V ....

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- Fall '05
- HUI
- Math, Linear Algebra, Algebra, Derivative, Vector Space, basis, dt