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Linear equations, solution sets and inner products

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1. The problem statement, all variables and given/known data

Let W be the subspace of R4 such that W is the solution set to the following system of equations:

x1-4x2+2x3-x4=0
3x1-13x2+7x3-2x4=0

Let U be subspace of R4 such that U is the set of vectors in R4 such the inner product <u,w>=0 for every w in W.

Find a 2 by 4 matrix B such that U is precisely the set of solutions in R4 of the homogenous system Bx=0.

2. Relevant equations



3. The attempt at a solution
Solving the system I got that W=(-3t+2s,t+s,s,t) where s and t are free variables. But I'm not sure how to get B from this information.
 
Assume that an element in U can be written as (x1,x2,x3,x4)

Now <u,w> = 0 for all w => you get an equation with t,s,x1,x2,x3,x4. This equation is true for all t and s, and hence the coefficient of t and s should be zero, right?

You therefore get two equations. Can you now construct B?
 

HallsofIvy

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1. The problem statement, all variables and given/known data

Let W be the subspace of R4 such that W is the solution set to the following system of equations:

x1-4x2+2x3-x4=0
3x1-13x2+7x3-2x4=0

Let U be subspace of R4 such that U is the set of vectors in R4 such the inner product <u,w>=0 for every w in W.

Find a 2 by 4 matrix B such that U is precisely the set of solutions in R4 of the homogenous system Bx=0.

2. Relevant equations



3. The attempt at a solution
Solving the system I got that W=(-3t+2s,t+s,s,t) where s and t are free variables. But I'm not sure how to get B from this information.
Assuming that is correct, we can write any vector in W (not "W= ", W is a set of vectors, not a vector.) as (-3t, t, 0, t)+ (2s, s, s, 0)= t(-3, 1, 0, 1)+ s(2, 1, 1, 0). That is, W is the subspace spanned by (-3, 1, 0, 1) and (2, 1, 1, 0).

Any vector having inner product (dot product) with such a vector (the orthogonal complement of W) must be of the form (a, b, c, d) such that (a, b, c, d).(-3, 1, 0, 1)= -3a+ b+ d= 0 and (a, b, c, d).(2, 1, 1, 0)= 2a+ b+ c= 0. Those two equations can be solved for two unknowns in terms of the other two and so written as a span of vectors like before.

For the last part, you want a matrix
[tex]\begin{bmatrix}a & b &c & d \\ e & f & g& h\end{bmatrix}[/tex]
such that
[tex]\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}u \\ v \\ w \\ x \end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}[/tex]
where [u v w x] is either of the two vectors you got spanning U.
 

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