I have a few eigenvector/value proofs and not really sure how to go about/start them, if someone could give me some ideas that would be great sorry if the notation in this post is bad, if you got any question about it i'll try to explain it a little better
show that if A is a nxn Hermitian matrix then all its eigenvalues are real
Show that any nonzero column vector v, v*v is a positive number
Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m, that correspond to distinct eigenvalues is independent
but X*X=sum(conj(x_i).x_i, i=1,..n) is real, so:
But A is hermitian so A*=A
hence s is real.
that correspond to distinct eigenvalues is independent
Suppose m=2, and that X1, and X2 are the e-vectors corresponding to
e-values lambda1, lambda2 and that lambda1 != lambda2, further suppose
that X1 and X2 are not linearly independent. Then there exist constants
a, b != 0 such that:
a.X1 + b.X2 = 0 ....(1)
A(a.X1+b.X2)=a.lambda1.X1 + b.lambda2.X2=0
from (1) we have a.X1=-b.X2, so:
but as b != 0, X2 !=0 and lambda1 != lambda2 this is
a contradiction, so we have the desired result is true
The above constitutes the base case for an inductive
proof that the result is true for any m<=n.
Suppose X(i), i=1..k, k<n-1 are linearly independent e-vectors
corresponding to distinct e-values lambda(i), i=1,..k. Further
suppose that X(i), i=1..k+1, k<n-1 are not linearly independent
e-vectors corresponding to distinct e-values lambda(i), i=1,..k+1.
Then there exist constants a(i), i=1,..k+1 not all zero such that:
sum(i=1,k+1) a(i).X(i) = 0,
a(k+1).X(k+1) = - sum(i=1,k) a(i).X(i)...(2)
Now multiply this by A:
lambda(k+1).a(k+1).X(k+1) = - sum(i=1,k) a(i).lambda(i).X(i).
Substitute from (2):
lambda(k+1).sum(i=1,k) a(i).lambda(i).X(i)= - sum(i=1,k) a(i).lambda(i).X(i)
but as lambda(k+1) != lambda(i), i=1,..k this implies that the X(i) i=1,..k
are not linearly independent - a contradiction.
This last result together with the base case proves by the desired
result by induction (it does not prove it true for all m in this case
as we have at most n distinct e-values)
Okay, I never tood a course in Linear Algebra (especially when it is already advanced) but I would like to make a recommendation that every finite dimensional vector space has a finite basis. Thus, maybe you use that here. Since the superset here is a basis (finite) then its subset surly contains a finite span set. Q.E.D.
considered simpler, but I don't see that what you are saying contributes to
that. The idea in the proof given is in fact pretty simple if a bit fussy to
That these e-vectors span a subspace is not in question, what is is what
is its dimension?
Also Adding QED to the end of a half formed idea does not make a proof.