Let A=(1,1,1,0),B=(−1,0,1,1,),C=(3,2,−1,1) and let D={Q∈R4|Q⊥A,Q⊥B,Q⊥C}. Convince me that D is a subspace of R4. Write D as span of a basis. Write D as a span of an orthogonal basis

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We will denote the scalar product as ⟨−,−⟩ Let v,w∈D and α,β∈R. Then ⟨αv+βw,A⟩=α⟨v,A⟩+β⟨w,A⟩=0 This is because, since v, w in D, we have that ⟨v,A⟩=0 and ⟨w,A⟩=0. So, αv+βw⊥A Similarly, we get that αv+βw⊥B and αv+βw⊥C Thus, αv+βw∈D. This proves that D is a subspace of R4 . Now let (x1,x2,x3,x4) in D. Then we have that ⟨(x1,x2,x3,x4),A⟩=0⇒x1+x2+x3=0 ⟨(x1,x2,x3,x4),B⟩=0⇒−x1+x2+x3=0 ⟨(x1,x2,x3,x4),C⟩=0⇒−3x1−2x2−x3+x4=0 Thus, we have a system of equations x1+x2+x3=0 −x1+x3+x4=0 −3x1−2x2−x3+x4=0 The matrix of this homogeneous system is [1110−1011−3−2−11] We will now use the following elementary row operations: 1. Interchange the ith and jth row: Ri↔Rj . 2, Multiply the eth row by a constant c≠0:cRi 3. Multiply the jth row by a constant c and add it to the ith row:  Rj+cRj [1110−1011−3−2−11] (R3+3R1)→[111001210121] (R3+3R1)→[10−1−101210121] (R3+3R1)→[10−1−101210000] This means that we have to have 2 parameters: x3=s,x4=t, and x1=s+t x2=−2s−t This means that

Let A = (1, 1, 1, 0), B = (-1, 0, 1, 1,), C = (3, 2, -1, 1) and let D = {Q in R^{4} | Q perp A, Q perp B, Q perp C}

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