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- Let C be the set of complex numbers. Define addition on C by (a + bi) + (c + di) = (a + c) + (b + d)i and define scalar multiplication by α(a + bi) = αa + αbi for all real numbers α. Show that C is a vector space with these operations.arrow_forwardDetermine whether the vectors u = 2- x + 3x2, v= 4 + x + 2x2, and w = 8-x+8x2 span P2 (the set of polynomials of degree 2 or less). If not find the space spanned by the given vectors.arrow_forwardIs the set of rational numbers with the usual definitions and scalar multiplication a vector space? Please show justification for answer and explain steps.arrow_forward
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- Find the transpose, conjugate, and adjoint of (i) Transpose is idempotent: (4¹) = A. (ii) Transpose respects addition: (A + B) = A¹+ B¹. (iii) Transpose respects scalar multiplication: (c. A)¹ = c A. These three operations are defined even when m‡n. The transpose and adjoint are both functions from CX to Cxm These operations satisfy the following properties for all c E C and for all A, B € CX". (iv) Conjugate is idempotent: A = A. (v) Conjugate respects addition: A 6-3i 0 1 B = A + B (vi) Conjugate respects scalar multiplication: C. A = c. A. (vii) Adjoint is idempotent: (4†)t = 4. (viii) Adjoint respects addition: (A + B) =A+B+. (ix) Adjoint relates to scalar multiplication: (CA)t = c · At 2 + 12i 5+2.1i 2+5i -19i 17 3-4.5iarrow_forwardRather than use the standard definitions of addition and scalar multiplication in R3, let these two operations be defined as shown below.With each of these new definitions, is R3 a vector space? Justify your answers.arrow_forwardFill up the blanks Let V be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on u = (u1, u2) and v = (v1, v2) : u +v= (1+ v1 – 1, u2 + V2 – 1), ku = (ku1, ku2) (a) Compute u + v and ku for u = (1,2), v = ( – 1,3) and k = 2. Solution: %3D %3D u + v= ku (b) Show that (0, 0) 0. Solution : We have (u1, u2) + (0,0) = ( ); thus u + (0,0) u for every || u in V. ) = 0. (c) Show that ( Solution: We have (u1, u2) + ( thus u + ( (u1 ) = (u1, u2); U2 = u for every u in V. %3D u such that u + (- u) = 0 (d) Show that Axiom 5 holds by producing an ordered pair for u = (u1, u2). Solution: If u = (u1, u2), then satisfies u + (- u) = ( - 6. ) = 0; thus Axiom 5 holds. (e) Find two vector space axioms that fail to hold. Solution: Axioms 7 fail to hold. For example: k(u + v) = k( a ) ( = (ku1, ku2) + (kv1, kv2) = ( %3D ku +k v || 6. k(u + v) ku +kv, k 1. Axioms 8 fail to hold. For example: = (k+1)(u1, u2) = ( ku + lu = (kuj, ku2) + (lu1,…arrow_forward
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