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}

snowlovelydayM 2021-01-13 Answered

Let A=(1,1,1,0),B=(1,0,1,1,),C=(3,2,1,1)
and let D={QR4|QA,QB,QC}.
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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Expert Answer

Dora
Answered 2021-01-14 Author has 98 answers

We will denote the scalar product as ,
Let v,wD and α,βR.
Then αv+βw,A=αv,A+βw,A=0
This is because, since v, w in D, we have that v,A=0 and w,A=0.
So, αv+βwA
Similarly, we get that
αv+βwB and αv+βwC
Thus, αv+βwD.
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=0x1+x2+x3=0
(x1,x2,x3,x4),B=0x1+x2+x3=0
(x1,x2,x3,x4),C=03x12x2x3+x4=0
Thus, we have a system of equations
x1+x2+x3=0
x1+x3+x4=0
3x12x2x3+x4=0
The matrix of this homogeneous system is [111010113211]
We will now use the following elementary row operations:
1. Interchange the ith and jth row: RiRj
. 2, Multiply the eth row by a constant c0:cRi
3. Multiply the jth row by a constant c and add it to the ith row:

Rj+cRj [111010113211]
(R3+3R1)[111001210121]
(R3+3R1)[101101210121]
(R3+3R1)[101101210000]
This means that we have to have 2 parameters: x3=s,x4=t, and
x1=s+t
x2=2st
This means that

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