Ortogonal subspace proof - Leon 5.2

[IntroAnalysis]

Homework Statement

Let S be a subspace of R³ spanned by the vectors x = (x₁, x₂, x₃)^T and y = (y₁, y₂, y₃)^T

Let A = (x₁ x₂ x₃ )
( y₁ y₂ y₃)

Show that S^$$\bot$$ = N(A).

Homework Equations

The Attempt at a Solution

Any hints?

 

[Mark44]

Homework Statement

Let S be a subspace of R³ spanned by the vectors x = (x₁, x₂, x₃)^T and y = (y₁, y₂, y₃)^T

Let A = (x₁ x₂ x₃ )
( y₁ y₂ y₃)

Show that S^$$\bot$$ = N(A).

Homework Equations

The Attempt at a Solution

Any hints?

Start with some definitions of the terms in this problem. Do you know what S^$$\bot$$ and N(A) mean?

How much do you understand about this problem? For example, what does it mean that S is spanned by those two vectors?

 

[IntroAnalysis]

The subspace S spanned by the two vectors means that any vector in S can be written as a linear combination of x and y. This means x and y are linearly independent.

S^$$\bot$$ is the orthogonal complement of S. It means the set w an element of R³ such that w^Ts= 0 for every s that's an element of S.

The null set N(A) is the set of all solutions to Ax = 0.

There exists z = (z₁, z₂, z₃) orthogonal to x and y.

(z1, z2, z3)^T* (x1, x2, x3) = 0 ; and (z1, z2, z3)^T* (y1, y2, y3) = 0

and the nullspace of N(A) = (x1 x2 x3) * (z1, z2, z3)^T = x1z1 + x2z2 +x3z3 = 0
(y1, y2, y3) * (z1, z2, z3)^T = y1z1 + y2z2 + y3z3 = 0

Therefore, the orthogonal complement of S = N(A)

Does that work? Thanks for your assistance!

 

[Mark44]

The subspace S spanned by the two vectors means that any vector in S can be written as a linear combination of x and y. This means x and y are linearly independent.

Not necessarily. For example, the vectors <1, 1> and <2, 2>} span a subspace of R² but they aren't linearly independent. This subspace has dimension 1.

S^$$\bot$$ is the orthogonal complement of S. It means the set w an element of R³ such that w^Ts= 0 for every s that's an element of S.

The null set N(A) is the set of all solutions to Ax = 0.

There exists z = (z₁, z₂, z₃) orthogonal to x and y.

(z1, z2, z3)^T* (x1, x2, x3) = 0 ; and (z1, z2, z3)^T* (y1, y2, y3) = 0

and the nullspace of N(A) = (x1 x2 x3) * (z1, z2, z3)^T = x1z1 + x2z2 +x3z3 = 0
(y1, y2, y3) * (z1, z2, z3)^T = y1z1 + y2z2 + y3z3 = 0

Therefore, the orthogonal complement of S = N(A)

Does that work? Thanks for your assistance!

You have the guts of the proof, but your presentation is on the clunky side. To show that two sets are equal, show that if u is in the first set, it must be in the second set. The show that if u is in the second set, it must also be in the first set.

So for your problem, assume that u is in N(A). You should be able to show that u is also in S$\bot$.
Now assume that u is in S$\bot$. Then show that u must also be in N(A).