lina-30-10-09-s1 - Linear Algebra & Geometry: solutions...

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Linear Algebra Geometry: solutions 4 1. We consider k v + w k 2 = ( v + w ) · ( v + w ) = v · v + 2 v · w + w · w = k v k 2 + k w k 2 + 2 v · w Now we use Cauchy Schwarz v · w ≤ | v · w | ≤ k v kk w k and obtain k v + w k 2 ≤ k v k 2 + k w k 2 + 2 k v kk w k = ( k v k + k w k ) 2 and taking the square root gives the triangle inequality. 2. Let us choose v = ( a 1 ,a 2 , ··· ,a N ) R N and w = (1 /N, 1 /N, ··· , 1 /N ) R N , then v · w = a 1 + a 2 + ··· + a N N , and k v k 2 = a 2 1 + a 2 2 + ··· + a 2 N , k w k 2 = 1 /N 2 + 1 /N 2 + ··· + 1 /N 2 = N/N 2 = 1 /N . So using the Cauchy Schwarz inequality in the form ( v · w ) 2 ≤ k v k 2 k w k 2 gives ± a 1 + a 2 + ··· + a N N ² 2 a 2 1 + a 2 2 + ··· + a 2 N N . 3. V is a subspace if (i) v + w V for all v , w V (ii) λ v V for all λ R and v V (a) The set of all vectors of the form x 0 0 , with x R . This is a subspace since: (i) x 0 0 + y 0 0 = x + y 0 0 is of the same form and (ii) λ x 0 0 = λx 0 0 , too. 1
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(b) The set of all vectors of the form x 1 1 , with x R . This is not a subspace since for instance 2
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lina-30-10-09-s1 - Linear Algebra & Geometry: solutions...

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