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