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Orthogonal complement (linear algebra)

  • Thread starter timon
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1. The problem statement

let [tex]\vec x[/tex] and [tex]\vec y[/tex] be linearly independent vectors in [tex]R^n[/tex] and let [tex]S=\text{span}(\vect x, \vect y).[/tex] Define the matrix [tex]A[/tex] as

[tex]A=\vec x \vec y^T + \vec y \vec x^T[/tex].​

Show that [tex]N(A)=S^{\bot}[/tex].

2.equations
I have a theorem that says[tex] N(A) = R(A^T)^{\bot}[/tex].
[tex]A[/tex] is symmetric; [tex] A = A^T[/tex].

3.Plan of attack
From the given above, it follows that if i can proof that [tex]S[/tex] is the orthogonal complement of [tex]R(A)[/tex], i'll be done. To do that, i'll have to show that all elements of [tex]S[/tex] are orthogonal to [tex]R(A)[/tex], and that any vector orthogonal to [tex]R(A)[/tex] is part of [tex]S[/tex].

Thus i want to show that
(I)the vectors [tex]\vec x[/tex] and [tex]\vec y[/tex] are orthogonal to any vector [tex]\vec z \in R(A)[/tex]
(II)any vector [tex]\vec k[/tex] that is orthogonal to all vectors [tex]\vec z \in R(A)[/tex] can be written as a linear combination of [tex]\vec x[/tex] and [tex]\vec y[/tex].

I'm really not seeing how to do this. Hope someone can help me out, or at least tell me if i'm on the right track. Cheers.
 
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Answers and Replies

  • #2
vela
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I think it would be easier just to show N(A)⊂S and S⊂N(A). Proving both directions is pretty straightforward.
 

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