using R language and provide R code Define column vectors v1 = (1, 1, 1)⊤, v2 = (1, 1, −1)⊤. (a) (10 pts) Compute ∥v1∥,∥v2∥, and v1 · v2. (b) (20 pts) Obtain vectors w1,w2, such that w1, w2 are orthogonal unit vectors, and span(v1, v2) = span(w1, w2). (c) (10 pts) Let w3 = (1/√2, −1/√2, 0)T . Check that w1 , w2 , w3 are mutually orthogonal unit vectors (i.e. they form an orthonormal basis of R3). (d) (10 pts) Define a 3×3 matrix A = [w1,w2,w3]. With the help of A, find linear coefficients x1, x2, x3, such that x1w1 + x2w2 + x3w3 = (3, 4, 5)⊤.
Question
using R language and provide R code Define column vectors v1 = (1, 1, 1)⊤, v2 = (1, 1, −1)⊤. (a) (10 pts) Compute ∥v1∥,∥v2∥, and v1 · v2. (b) (20 pts) Obtain vectors w1,w2, such that w1, w2 are orthogonal unit vectors, and span(v1, v2) = span(w1, w2). (c) (10 pts) Let w3 = (1/√2, −1/√2, 0)T . Check that w1 , w2 , w3 are mutually orthogonal unit vectors (i.e. they form an orthonormal basis of R3). (d) (10 pts) Define a 3×3 matrix A = [w1,w2,w3]. With the help of A, find linear coefficients x1, x2, x3, such that x1w1 + x2w2 + x3w3 = (3, 4, 5)⊤.
Solution
Sure, here is the R code for each part of your question:
(a) Compute ∥v1∥,∥v2∥, and v1 · v2.
# Define the vectors
v1 <- c(1, 1, 1)
v2 <- c(1, 1, -1)
# Compute the norms
norm_v1 <- sqrt(sum(v1^2))
norm_v2 <- sqrt(sum(v2^2))
# Compute the dot product
dot_product <- sum(v1 * v2)
(b) Obtain vectors w1,w2, such that w1, w2 are orthogonal unit vectors, and span(v1, v2) = span(w1, w2).
# Define w1 as v1 normalized
w1 <- v1 / norm_v1
# Define w2 as the orthogonal projection of v2 onto w1, normalized
w2 <- (v2 - sum(v2 * w1) * w1) / sqrt(sum((v2 - sum(v2 * w1) * w1)^2))
(c) Let w3 = (1/√2, −1/√2, 0)T . Check that w1 , w2 , w3 are mutually orthogonal unit vectors (i.e. they form an orthonormal basis of R3).
# Define w3
w3 <- c(1/sqrt(2), -1/sqrt(2), 0)
# Check orthogonality
orthogonal <- all(abs(c(sum(w1 * w2), sum(w1 * w3), sum(w2 * w3))) < 1e-10)
# Check unit vectors
unit_vectors <- all(abs(c(sqrt(sum(w1^2)), sqrt(sum(w2^2)), sqrt(sum(w3^2))) - 1) < 1e-10)
(d) Define a 3×3 matrix A = [w1,w2,w3]. With the help of A, find linear coefficients x1, x2, x3, such that x1w1 + x2w2 + x3w3 = (3, 4, 5)⊤.
# Define the matrix A
A <- cbind(w1, w2, w3)
# Define the vector (3, 4, 5)
v <- c(3, 4, 5)
# Solve the linear system
x <- solve(A, v)
The vector x contains the coefficients x1, x2, x3.
Similar Questions
YouLet v1=[-1 4 0 0 0 -1], v2=[-5 0 2 -1 0 -2], v3=[5 0 0 3 3 5]Use the Gram-Schmidt procedure to produce an orthogonal set with the same span. what is u1, <u1, v2>, <u1, u1>, u2, <u1, v3>, <u2, v3>, <u2, u2>, and u3.
Let W1 = L{(1, 1, 0), (−1, 1, 0)} and W2 = L{(1, 0, 2), (−1, 0, 4)}. Show that W1 +W2 = R3. Give an example of a vector v ∈ R3 such that v can be written in twodifferent ways in the form v = v1 + v2, where v1 ∈ W1, v2 ∈ W2
Use the inner product u, v = 2u1v1 + u2v2 in R2 and the Gram-Schmidt orthonormalization process to transform {(2, 1), (−2, −9)} into an orthonormal basis. (Use the vectors in the order in which they are given.)
Let u, v and w be vectors in R3. Which of the following calculations is possible? (a) u + (v · w) (b) u · (v × w) (c) (u · v) × w (d) u × (v · w)
Let's solve the given homework question step by step: (a) Write the vector \( w \) as a linear combination of \( u \) and \( v \). Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] We want to find scalars \( a \) and \( b \) such that \( w = au + bv \). This gives us the following system of equations: \[ a(1) + b(2) = -5 \] \[ a(1) + b(-1) = 1 \] Solving this system, we get: \[ a + 2b = -5 \] \[ a - b = 1 \] Adding the two equations, we get: \[ 2a + b = -4 \] \[ a = -\frac{5}{2} \] Substituting \( a \) back into the second equation: \[ -\frac{5}{2} - b = 1 \] \[ b = -\frac{7}{2} \] So, \( w \) can be written as a linear combination of \( u \) and \( v \) with \( a = -\frac{5}{2} \) and \( b = -\frac{7}{2} \): \[ w = -\frac{5}{2}u - \frac{7}{2}v \] (b) Are the vectors \( u \), \( v \), and \( w \) linearly independent? Vectors are linearly independent if the only solution to the equation \( \alpha u + \beta v + \gamma w = 0 \) is \( \alpha = \beta = \gamma = 0 \). Since we were able to express \( w \) as a linear combination of \( u \) and \( v \), this means that there exists a non-trivial solution to the equation \( \alpha u + \beta v + \gamma w = 0 \) (specifically, \( \alpha = -\frac{5}{2} \), \( \beta = -\frac{7}{2} \), and \( \gamma = 1 \)). Therefore, \( u \), \( v \), and \( w \) are not linearly independent. (c) Show that span(\( \{u, v, w\} \)) = \( \mathbb{R}^2 \) by showing that an arbitrary vector in \( \mathbb{R}^2 \) can always be expressed as a linear combination of \( u \), \( v \), \( w \). Since \( u \) and \( v \) are linearly independent (you can verify this by checking that the matrix formed by \( u \) and \( v \) has a non-zero determinant), any vector in \( \mathbb{R}^2 \) can be expressed as a linear combination of \( u \) and \( v \). Therefore, the span of \( \{u, v\} \) is \( \mathbb{R}^2 \), and since \( w \) is also a linear combination of \( u \) and \( v \), the span of \( \{u, v, w\} \) is still \( \mathbb{R}^2 \). (d) What is the dimension of the vector space span(\( \{u, v, w\} \))? The dimension of a vector space is the number of vectors in its basis. Since \( u \) and \( v \) are linearly independent and span \( \mathbb{R}^2 \), they form a basis for \( \mathbb{R}^2 \). The vector \( w \) does not add any new dimension because it is a linear combination of \( u \) and \( v \). Therefore, the dimension of the vector space span(\( \{u, v, w\} \)) is 2, which is the same as the dimension of \( \mathbb{R}^2 \).
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