Thread: Real Symmetric Matrix

Consider the following real symmetric matrix:
0 1 -1
1 0 1
-1 1 0

it has eigenvalues t1=-2,t2=1,t3=1
with corresponding eigenvectors:
v1=(1,-1,1)^T
v2=(1,0,-1)^T
v3=(0,1,1)^T

However the inner product of v2 and v3, <v2,v3>=-1
Can anyone explain why this isn't 0?

Originally Posted by sym0110

Consider the following real symmetric matrix:
0 1 -1
1 0 1
-1 1 0

it has eigenvalues t1=-2,t2=1,t3=1
with corresponding eigenvectors:
v1=(1,-1,1)^T
v2=(1,0,-1)^T
v3=(0,1,1)^T

However the inner product of v2 and v3, <v2,v3>=-1
Can anyone explain why this isn't 0?

You seem to believe that eigenvectors of symmetric matrices are perpendicular to each other: this is false, and as an easy counter-example take the identity matrix for which ALL the non-zero vectors are its eigenvectors. Now just take a non-orthogonal basis...

Of course, your matrix is also a counterexample.

Tonio

Thx for the reply.
However I read in books that eigenvalues of symmetric matrices are orthogonal and their inner products are 0... moreover a matrix is symmetric iff its eigenvectors form an orthogonal basis...

Originally Posted by sym0110

Thx for the reply.
However I read in books that eigenvalues of symmetric matrices are orthogonal and their inner products are 0... moreover a matrix is symmetric iff its eigenvectors form an orthogonal basis...

Oh, hollie mollie! In what book did you read that nonsense?? Tell me please so that I can go and burn up all the copies I can find.

Tonio

Eigenvectors of symmetric matrices are orthogonal if and only if they are from different eigenspaces. (So if labda1 is not equal to labda 2)

Proof is easy by using that (A^t)v1=Av1 and just expending it on both sides:

labda2*v2*v1=Av2*v1=((v2^t)*(A^t))v1=(v2^t)*Av1=(v 2^t)*labda1*v1=labda1*v2*v1.

If labda 1 is not equal to labda 2, then this can only hold when v2*v1=0, hence are orthogonal.

In the counterexamples these were eigenvectors from the same eigenspace.