Title Linear combination and spanning set_02.pdf ✅

Linear Combination and Spanning Sets Page | 17 Definition 10.2.1. A vector v in a vector space V is called a linear combination of the vectors vଵ, vଶ,..., v௡ in V if v can be written in the form vൌcଵvଵ ൅ cଶvଶ ൅⋯൅c௡v௡ where cଵ, cଶ,..., c௡ are scalars. Example 10.2.1. If v ൌ ൥ 1 1 1 ൩, then v is a linear combination of vectors in the set S where S ൌ ⎩ ⎪ ⎨ ⎪ ⎧ ൥ 1 2 3 ൩ ถvభ , ൥ 0 1 2 ൩ ถvమ , ൥ െ1 0 1 ൩ ถvయ ⎭ ⎪ ⎬ ⎪ ⎫ . Solution. To see if v is a linear combination of vectors in S, consider the following equation: cଵvଵ ൅ cଶvଶ ൅ cଷvଷ ൌ v cଵ ൥ 1 2 3 ൩ ൅ cଶ ൥ 0 1 2 ൩ ൅ cଷ ൥ െ1 0 1 ൩ ൌ ൥ 1 1 1 ൩ ൥ cଵ െ cଷ 2cଵ ൅ cଶ 3cଵ ൅ 2cଶ ൅ cଷ ൩ ൌ ൥ 1 1 1 ൩ The augmented matrix becomes ൥ 1 2 3 0 1 2 െ1 0 1 อ 1 1 1 ൩ ሾvଵ vଶ vଷ | vሿ Row reduction gives ൥ 1 2 3 0 1 2 െ1 0 1 อ 1 1 1 ൩ ିோభାோమ→ோమ ିଷோభାோయ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 0 1 2 െ1 2 4 อ 1 െ1 െ2 ൩ ିଶோమାோయ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 0 1 0 െ1 2 0 อ 1 െ1 0 ൩ The reduced row echelon form shows that cଷ is free. Therefore, if cଷ ൌ 1, then cଵ ൌ 2, cଶ ൌ െ3. Therefore, v ൌ 2vଵ െ 3vଶ ൅ vଷ.
Linear Combination and Spanning Sets Page | 18 Remark This linear combination is not unique. For example, if cଷ ൌ 0, then cଵ ൌ 1, cଶ ൌ െ1. This gives vൌvଵ െ vଶ. Example 10.2.2. Show that vൌቂ0 8 2 1ቃ is a linear combination of the vectors in the set S ൌ ൞ቂ 0 2 1 0ቃ ᇣᇤᇥ vభ , ቂ െ1 3 1 2ቃ ᇣᇧᇤᇧᇥ vమ , ቂ െ2 0 1 3ቃ ᇣᇧᇤᇧᇥ vయ ൢ. Solution. We want to solve the vector equation for cଵ, cଶ and cଷ. cଵvଵ ൅ cଶvଶ ൅ cଷvଷ ൌ v cଵ ቂ 0 2 1 0ቃ൅cଶ ቂ െ1 3 1 2ቃ൅cଷ ቂ െ2 0 1 3ቃൌቂ0 8 2 1ቃ ൤ െcଶ െ 2cଷ 2cଵ ൅ 3cଶ cଵ ൅ cଶ ൅ cଷ 2cଶ ൅ 3cଷ ൨ൌቂ0 8 2 1ቃ The system becomes െcଶ െ 2cଷ ൌ 0 2cଵ ൅ 3cଶ ൌ 8 cଵ ൅ cଶ ൅ cଷ ൌ 2 2cଶ ൅ 3cଷ ൌ 1 Therefore, the reduced row echelon form may be obtained as follows: ൦ 0 2 1 0 െ1 3 1 2 െ2 0 1 3 ተ 0 8 2 1 ൪ ோభ↔ோయ ሱ⎯⎯⎯ሮ ൦ 1 2 0 0 1 3 െ1 2 1 0 െ2 3 ተ 2 8 0 1 ൪ ିଶோభାோమ→ோమ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 1 1 െ1 2 1 െ2 െ2 3 ተ 2 4 0 1 ൪ ோమାோయ→ோయ ିଶோమାோర→ோర ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 1 1 0 0 1 െ2 െ4 7 ተ 2 4 4 െ7 ൪ ି ଵ ସ ோయ→ோయ ଵ ସ ோర→ோర ሱ⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 1 1 0 0 1 െ2 1 1 ተ 2 4 െ1 െ1 ൪ ିோయାோర→ோర ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 1 1 0 0 1 െ2 1 0 ተ 2 4 െ1 0 ൪
Linear Combination and Spanning Sets Page | 19 ିோమାோభ→ோభ ሱ⎯⎯⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 0 1 0 0 3 െ2 1 0 ተ െ2 4 െ1 0 ൪ ଶோయାோమ→ோమ ିଷோయାோభ→ோభ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൦ 1 0 0 0 0 1 0 0 0 0 1 0 ተ 1 2 െ1 0 ൪ The solution is given by cଵ ൌ 1, cଶ ൌ 2, cଷ ൌ െ1. ∴vൌvଵ ൅ 2vଶ െ vଷ. Non‐Example. The vector v ൌ ൥ 1 െ2 2 ൩ is not a linear combination of vectors in the set S of Example 10.2.1. To see this—we consider the following equation: cଵvଵ ൅ cଶvଶ ൅ cଷvଷ ൌ v cଵ ൥ 1 2 3 ൩ ൅ cଶ ൥ 0 1 2 ൩ ൅ cଷ ൥ െ1 0 1 ൩ ൌ ൥ 1 െ2 2 ൩ ሺ1ሻ As in the previous example, the augmented matrix becomes ൥ 1 2 3 0 1 2 െ1 0 1 อ 1 െ2 2 ൩ ሾvଵ vଶ vଷ | vሿ Row reduction gives ൥ 1 2 3 0 1 2 െ1 0 1 อ 1 െ2 2 ൩ ିோభାோమ→ோమ ିଷோభାோయ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 0 1 2 െ1 2 4 อ 1 െ4 െ5 ൩ ିଶோమାோయ→ோయ ሱ⎯⎯⎯⎯⎯⎯⎯⎯ሮ ൥ 1 0 0 0 1 0 െ1 2 0 อ 1 െ1 3 ൩ The last row shows that the eq. (1) has no solution. Therefore, there exist no scalars cଵ, cଶ and cଷ such that the eq. (1) holds true.