Let B = {v1, V2, V3, V4} be an ordered basis for a vector space V. Let Vị = span{V1, V2} and 1 0 1 1 1 0 1 V2 = span{v3, v4} . If MB(T) , prove that T(V1)NT(V2) = {0} . Note: 1 1 0 0 0 1 T(V1) = {T(x) : x € Vị} and T(V2) = {T(x) : x e V2} .

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A: We have to find out solutions

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A: ANSWER:

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A: Note:  As you requested only Part C(I) so I answered here only that.

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A: value of k is 0 see calculation below

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Q: Let V be an n-dimensional vector space with an ordered basis β. Define T: V →Fn by T(x) = [x]β.…

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A: We have to find p(x) and q(x).

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Q: -2 0 T 0 0 0 1 -10x3 +4x2 -4x+2 T 1 1 0 0 = 0 T 11 11 -5x3 +2x2-2x+1 T -5x3 +2x2-2x+1 = = 1 2 Give a…

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A: First we prove it and next we illustrated by examples.

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A: Let ℝn be a vector space. Let W be a subspace of ℝn. Then,  is W={v1→,......,vp→}a basis. Definition…

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Q: 5. Let V be a vector space, and let x1, X2,.. . , Xk E V. 5(a) Prove that if either x1 = 0 or x;+1 E…

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A: 2. (a) Given list is p0x, p1x, p2x, ..., pnx, ..., where p0x=1,    p1x=1+x,    ......,…

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A: Concept: A branch of mathematics which deals with symbols and the rules for manipulating those…

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A: Dot product of the two vectors.

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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.