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Let be U = Span{[3,0,1]} a subspace of W=Span{[1,0,0],[1,0,1]}, and the space V the orthogonal complement of U. Answer these two following questions:1. Give a basis v=[ , , ] for V,2. Give dim V = .(If your result is not an integer, then please give your result with 2 decimals (rounded - not cut-off; use a dot as a decimal separator))

Question

Let be U = Span{[3,0,1]} a subspace of W=Span{[1,0,0],[1,0,1]}, and the space V the orthogonal complement of U. Answer these two following questions:1. Give a basis v=[ , , ] for V,2. Give dim V = .(If your result is not an integer, then please give your result with 2 decimals (rounded - not cut-off; use a dot as a decimal separator))

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Solution

  1. The orthogonal complement of U in W is the set of all vectors in W that are orthogonal to every vector in U. Since U is spanned by the vector [3,0,1], we need to find a vector in W that is orthogonal to [3,0,1].

The vectors in W are linear combinations of [1,0,0] and [1,0,1]. Let's denote a vector in W as [a,b,c] = x[1,0,0] + y[1,0,1] = [x+y,0,y]. This vector is orthogonal to [3,0,1] if their dot product is zero, i.e., (x+y)3 + 00 + y*1 = 0. Solving this equation gives y = -3x.

Substituting y = -3x into [x+y,0,y] gives the vector [x-3x,0,-3x] = [-2x,0,-3x]. For this vector to be non-zero, we can choose x = 1/2, which gives the vector [-1,0,-3/2]. Therefore, a basis for V is v = [-1,0,-3/2].

  1. The dimension of a vector space is the number of vectors in its basis. Since V has one vector in its basis, the dimension of V is 1.

This problem has been solved

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