Can someone help me about skew symmetric?

JerryKelly
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Let A be an nxn skew symmetric mx.(A^T=-A).
i) Show that if X is a vector in R^n then (X,AX)=0
ii) Show that 0 is the only possible eigenvalue of A
iii)Show that A^2 is symmetric
iv)Show that every eigenvalue of A^2 is nonpositive.
v)Show that if X is an eigenvector of A^2 , then so is AX
vi)With X as in v), show that the subspace W spanned by X and AX is an A-subspace.
vii)Assuming AX not= 0. let U=X/||X||, V=AX/||AX||. Show that AU=(AU,V)V and that AV=-(AU,V)U.
ix) Show if U,V,U3 , ...,Un is an orthonormal basis of R^n, then the mx of a reletive to this basis has the form (0,b,0,0...0;-b,0,0,0...0;0,0,0,a,0...0;0,0,-a,0,0...0;C some format to the end), where C is skew symmetric (n-2)x(n-2) mx.


Thanks!
 
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This smells like homework. What have you tried on these questions so far?
 
i) follows from the definition of transpose and the properties of inner products, what do you know about them?
 
Yes,it is homework question. So far, I just have a idea for part iv), and I have no idea for the rest of them. For part iv), i can use (AX,X)=(X,-AX)=-A(X,X). Except this one, I have no idea how to do the rest of them. Could you give me some help,please? Thanks!
Galileo said:
This smells like homework. What have you tried on these questions so far?
 
I think you need to relearn some basics, or correct some typos.

How can (x,-Ax)=-A(x,x)?

The left hand side is a scalar, the right hand side is a matrix (times a scalar)

Remember the definition of transpose is made to satisfy

(Mx,y)=(x,M^Ty)

and that pretty much is all you need, in fact it solves pretty much everything (along with the definition of eigenvalue/vector)
 
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Could you solve the last two parts by using this definition for me,please? I have no any idea for the last two parts. Thanks!
matt grime said:
I think you need to relearn some basics, or correct some typos.

How can (x,-Ax)=-A(x,x)?

The left hand side is a scalar, the right hand side is a matrix (times a scalar)

Remember the definition of transpose is made to satisfy

(Mx,y)=(x,M^Ty)

and that pretty much is all you need, in fact it solves pretty much everything (along with the definition of eigenvalue/vector)
 
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