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X and y lin independent in Rn A=xy^T+yx^T

  • Thread starter Keesjan
  • Start date
  • #1
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Homework Statement


Well as the title describes, x and y linear independent in R^n S is a subspace in Rn spanned by x,y i.e S= span(x,y)
define the matrix A as A=xy^T+yx^T

This is actually a 3 part question

1 show that A is symmetrical
2 show that N(A)=S(Perpendicular)
3 show that the rank of A is 2

Homework Equations



The Attempt at a Solution



I think i got the first one
A^T= (xy^T+yx^T)^T = (y^T)^Tx^T+(x^T)^Ty^T= yx^T+xy^T=A QED

Dimension is abbreviated as Dim
Rowspace abbreviated as R(*)
Column space will be abbreviated as C(*)
and perpendicular as perp

Now the second one i think has something to do with the N(A)=C(A^T)^perp.
And because A^T=A now N(A)=C(A)^perp
Span of the DimC(A) is span of S because x,y are linear independent?
And if DimC(A)=Span(x,y) then DimC(A)=S

So filling it all in N(A)=S^perp QED??

is this correct?

Last question

Since x and y are linear independent they form the basis of A. So DimC(A)=2
And since the DimC(A)=DimR(A) and the rank=DimR so Rank(A)=2

Any feed back would be nice.

Peace

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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To show that N(A)= S(perpendicular) You need to show that A(u)= 0 (so u in in N(A), the null space of A) if and only if <u, v>= 0 for all v in S.
 
  • #3
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To show that N(A)= S(perpendicular) You need to show that A(u)= 0 (so u in in N(A), the null space of A) if and only if <u, v>= 0 for all v in S.
What do you mean with <u,v>=0 for all v in S
Maby i dont under stand the notation but could you explain a little more?

Thanks
 

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