Linear equations, solution sets and inner products

In summary, find a 2 by 4 matrix B such that U is precisely the set of solutions in R4 of the homogenous system Bx=0.
  • #1
reb659
64
0

Homework Statement



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.

Homework Equations





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.
 
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  • #2
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?
 
  • #3
reb659 said:

Homework Statement



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.

Homework Equations





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.
 

Related to Linear equations, solution sets and inner products

1. What is a linear equation?

A linear equation is an algebraic equation that contains only variables with degree 1 and constants. It can be written in the form of y = mx + b, where m is the slope and b is the y-intercept. The graph of a linear equation is a straight line.

2. How do you solve a linear equation?

To solve a linear equation, you must isolate the variable on one side of the equation by using inverse operations. This means that you must perform the opposite operation to both sides of the equation to cancel out the variable's coefficient. The resulting value of the variable is the solution to the equation.

3. What is a solution set?

A solution set is a set of values that make an equation true. In the case of a linear equation, it is the set of values for the variable that satisfies the equation. The solution set can be a single value, multiple values, or even infinite values depending on the equation.

4. What is an inner product?

An inner product is a mathematical operation that takes two vectors as input and produces a scalar value as output. It is often denoted by and is used to calculate the angle between two vectors or to project one vector onto another.

5. How are linear equations and inner products related?

Linear equations and inner products are related through the concept of orthogonality. Two vectors are said to be orthogonal if their inner product is equal to 0. In the case of linear equations, the solution set represents the values of the variables that make the equations orthogonal to each other.

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