- #1
megaman
- 1
- 0
I am looking for a rather simple proof that if the matrix A has eigenvalues>0, then (x^t)Ax>0 for any vector x not 0.
My first tought was if the eigenvalues are bigger than 0, then (x^t)Ax=(x^t)"eigenvaulue"x="eigenvalue"(x^t)x>0, if x is nonzero and eigenvalues is bigger than 0.
Is this proof good enough? I am a little unsure, becuase I think I have not proven it for all x, just the eigenvectors.
My first tought was if the eigenvalues are bigger than 0, then (x^t)Ax=(x^t)"eigenvaulue"x="eigenvalue"(x^t)x>0, if x is nonzero and eigenvalues is bigger than 0.
Is this proof good enough? I am a little unsure, becuase I think I have not proven it for all x, just the eigenvectors.