# Eigenvectors/value proofs

• Oct 2nd 2006, 02:24 PM
action259
Eigenvectors/value proofs
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
• Oct 2nd 2006, 04:24 PM
ThePerfectHacker
Quote:

Originally Posted by action259

Show that any nonzero column vector v, v*v is a positive number

If you have,
v=[a1 a2 a3 .... an]
Given that it is nonzero, i.e. a1,a2,... not equal to zero.
Then,
v*v=(a1)^2+(a2)^2+...+(an)^2 sum of square is positive.
All of them cannot be zero.
• Oct 2nd 2006, 08:50 PM
CaptainBlack
Quote:

Originally Posted by action259
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

Let A be hermitian, and s be an eigen-value corresponding to eigen vector X.

Then:

AX=sX

So:

X*AX=sX*X,

but X*X=sum(conj(x_i).x_i, i=1,..n) is real, so:

X*AX/(X*X)=s

and:

conj(X*AX/(X*X))=conj(X*AX)/(X*X)=(X*A*X)/(X*X)=conj(s).

But A is hermitian so A*=A

so:

conj(s)=(X*AX)/(X*X)=s,

hence s is real.

RonL
• Oct 3rd 2006, 12:16 AM
CaptainBlack
Quote:

Originally Posted by action259
Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m, that correspond to distinct eigenvalues is independent

Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m,
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)

Now:

A(a.X1+b.X2)=a.lambda1.X1 + b.lambda2.X2=0

from (1) we have a.X1=-b.X2, so:

(lambda1-lambda2).b.X2=0,

but as b != 0, X2 !=0 and lambda1 != lambda2 this is
a contradiction, so we have the desired result is true
when m=2.

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,

so:

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)

or:

sum(i=1,k) a(i).(lambda(i)-lambda(k+1)).X(i)=0.

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)

RonL
• Oct 3rd 2006, 08:22 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m,
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)

Now:

A(a.X1+b.X2)=a.lambda1.X1 + b.lambda2.X2=0

from (1) we have a.X1=-b.X2, so:

(lambda1-lambda2).b.X2=0,

but as b != 0, X2 !=0 and lambda1 != lambda2 this is
a contradiction, so we have the desired result is true
when m=2.

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,

so:

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)

or:

sum(i=1,k) a(i).(lambda(i)-lambda(k+1)).X(i)=0.

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)

RonL

And think there is an easier way.
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.
• Oct 3rd 2006, 08:48 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
And think there is an easier way.
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.

I'm pretty sure that there is a proof that on some measure might be
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
prove.

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.

RonL