Let vector y= [ -1 3 5] and vector u= [-5 6 -7]. Write vector y as the sum of two orthogonal vectors, what are the vector x1 in span {u} and vector x2 orthogonal to vector u.

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 x=[1, 0, -1], and let U be the subspace spanned by the orthogonal set {[2, 1, -1],[1,1,3]}. Vector x can be written as x=x1+x2, where x1 is in U and x2 is orthogonal to U. Find the first coordinate of x2.Select one:a. 4/11b. 6/11c. None of the other choices is correctd. 5/11e. 3/11

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 \( \mathbb{R}^2 \).

Let  and  be two vectors.

Let x and y be two vectors from Rn. Show that x − y andx + y are orthogonal if and only if ∥x∥ = ∥y∥