Solved 1 Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Chegg There are 3 steps to solve this one. 1. let v1 = (4,0,2,4),v2 =(3,0,3,0), and v3 = (1,4,3,5). let v =span{v1,v2,v3}. (a) (5 points) apply the gram schmidt process to transform {v1,v2,v3} into an orthonormal basis for v. (b) (5 points) let u= (0,−5,16,13). find the orthogonal projection of u onto v. not the question you’re looking for?. Let {v1, v2, . . . , vk} be a set of at least two vectors in a vector space v . if one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors and the linear span of the resulting set of vectors will be the same as the linear span of v1, v2, . . .
Solved 1 Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Chegg V2 · v2 = (3, 0, 3, 0) · (3, 0, 3, 0) = 3^2 0^2 3^2 0^2 = 18. v3 · v3 = (1, 4, 3, 5) · (1, 4, 3, 5) = 1^2 4^2 3^2 5^2 = 51. now let's calculate the dot products between u and each of the vectors: u · v1 = (0, 5, 16, 13) · (4, 0, 2, 4) = 0 0 32 52 = 84. u · v2 = (0, 5, 16, 13) · (3, 0, 3, 0) = 0 0 48 0 = 48. u. Given points p1 = [ 1 5 ] , p2 = [ 3 1 ] , and p3 = [ 1 2] in r^2, let s = conv {p1, p2, p3}. for each linear functional f , find the maximum value m of f on the set s, and find all points x in s at which f (x) = m. a. f1 (x1, x2) = x1 x2 b. f2 (x1, x2) = 3×1 x2 c. f3 (x1, x2) = x1 2×2. Below i've provided a few methods of determining whether or not they are linearly independent. method #1. ⎡⎣⎢1 1 2 1 0 1 2 2 3⎤⎦⎥ [1 1 2 1 0 2 2 1 3] now use gauss jordan elimination to determine the rank of the matrix. if it is 3 3 (the number of columns), then the columns are linearly independent. method #2. construct the same matrix as above. V1=(1,6,4), v2=(2,4, 1), v3=( 1,2,5) and w1=(1, 2, 5), w2=(0,8,9). prove that span(v1,v2,v3)=span(w1,w2). try to solve aw1 bw2 = v1, cw1 dw2 = v2, and ew1 fw2 = v3. i.e. a(1, 2, 5) b(0,8,9)=( 1,6,4) (1) c(1, 2, 5) d(0,8,9)=(2,4, 1) (2) e(1, 2, 5) f(0,8,9)=( 1,2,5) (3) (1) becomes: a 0=1, 2a 8b = 6, 5a 9b = 4. so,a= 1, b=1.
Solved 1 Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Chegg Below i've provided a few methods of determining whether or not they are linearly independent. method #1. ⎡⎣⎢1 1 2 1 0 1 2 2 3⎤⎦⎥ [1 1 2 1 0 2 2 1 3] now use gauss jordan elimination to determine the rank of the matrix. if it is 3 3 (the number of columns), then the columns are linearly independent. method #2. construct the same matrix as above. V1=(1,6,4), v2=(2,4, 1), v3=( 1,2,5) and w1=(1, 2, 5), w2=(0,8,9). prove that span(v1,v2,v3)=span(w1,w2). try to solve aw1 bw2 = v1, cw1 dw2 = v2, and ew1 fw2 = v3. i.e. a(1, 2, 5) b(0,8,9)=( 1,6,4) (1) c(1, 2, 5) d(0,8,9)=(2,4, 1) (2) e(1, 2, 5) f(0,8,9)=( 1,2,5) (3) (1) becomes: a 0=1, 2a 8b = 6, 5a 9b = 4. so,a= 1, b=1. There are 2 steps to solve this one. in this process we 1. let v1 = (4,0,2,4),v2 =(3,0,3,0), and v3 = (1,4,3,5). let v =span{v1,v2,v3}. (a) (5 points) apply the gram schmidt process to transform {v1,v2,v3} into an orthonormal basis for v. (b) (5 points) let u= (0,−5,16,13). find the orthogonal projection of u onto v. To determine if the vectors v1 = [0, 0, 3], v2 = [0, 3, 9], and v3 = [4, 2, 6] span r³, we need to check if they are linearly independent. to do this, we can set up a matrix with these vectors as columns and attempt to reduce it to its row echelon form. 3. given the subset s = {(1,2,1), (2,3,3), (−3,5,−14), (0,1,−1)} of r3, deter mine whether the set s spans r3. if not, find a vector which is not in the subspace of r3 spanned by s. solution: an arbitrary vector (a,b,c) ∈ r3 is in the subspace of r3 spanned s if and only if there are real numbers c 1,c 2,c 3,c 4 such that c 1(1,2,1) c. Let v1, v2, v3 be in r^n, and let a = [v1 v2 v3]. determine whether each if then statement is true or false. if true, explain why. if false, provide a specific numerical example of vectors v1, v2, v3 for which the “if” holds but the “then” does not. if {v1, v2, v3} is linearly independent, then the only solution to ax = 0 is x = 0.
Solved 1 Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Chegg There are 2 steps to solve this one. in this process we 1. let v1 = (4,0,2,4),v2 =(3,0,3,0), and v3 = (1,4,3,5). let v =span{v1,v2,v3}. (a) (5 points) apply the gram schmidt process to transform {v1,v2,v3} into an orthonormal basis for v. (b) (5 points) let u= (0,−5,16,13). find the orthogonal projection of u onto v. To determine if the vectors v1 = [0, 0, 3], v2 = [0, 3, 9], and v3 = [4, 2, 6] span r³, we need to check if they are linearly independent. to do this, we can set up a matrix with these vectors as columns and attempt to reduce it to its row echelon form. 3. given the subset s = {(1,2,1), (2,3,3), (−3,5,−14), (0,1,−1)} of r3, deter mine whether the set s spans r3. if not, find a vector which is not in the subspace of r3 spanned by s. solution: an arbitrary vector (a,b,c) ∈ r3 is in the subspace of r3 spanned s if and only if there are real numbers c 1,c 2,c 3,c 4 such that c 1(1,2,1) c. Let v1, v2, v3 be in r^n, and let a = [v1 v2 v3]. determine whether each if then statement is true or false. if true, explain why. if false, provide a specific numerical example of vectors v1, v2, v3 for which the “if” holds but the “then” does not. if {v1, v2, v3} is linearly independent, then the only solution to ax = 0 is x = 0.
Solved 1 Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Chegg 3. given the subset s = {(1,2,1), (2,3,3), (−3,5,−14), (0,1,−1)} of r3, deter mine whether the set s spans r3. if not, find a vector which is not in the subspace of r3 spanned by s. solution: an arbitrary vector (a,b,c) ∈ r3 is in the subspace of r3 spanned s if and only if there are real numbers c 1,c 2,c 3,c 4 such that c 1(1,2,1) c. Let v1, v2, v3 be in r^n, and let a = [v1 v2 v3]. determine whether each if then statement is true or false. if true, explain why. if false, provide a specific numerical example of vectors v1, v2, v3 for which the “if” holds but the “then” does not. if {v1, v2, v3} is linearly independent, then the only solution to ax = 0 is x = 0.
Solved Let V1 4 0 2 4 V2 3 0 3 0 And V3 1 4 3 5 Let Chegg