Please see the attached file for the fully formatted problems.
a. quote a theorem which guarantees that there exists an orthogonal basis for (with standard inner product) made up of eigenvectors of matrix
b. Find such a basis .
c. Represent the quadratic form by a symmetric matrix. Is Q positive definite? Justify your answer.
a. I am not sure this is right but here is theorem that we learned in class.
Let T be a linear operator on a finite-dimensional real inner product space V. Then T is self-adjoint if and only inf there exists an orthonormal basis for V consisting of eigenvectors of T.
( we know that A is self-adjoint if )
Since A is self-adjoint I guess I can use theorem 1???
What is the connection between T and A??
b. To find a basis made up of eigenvectors of matrix A, first of all, we know that we need to find the eigenvalues. Using det(tI-A) I found that and . (please check that it is correct).
For , I found eigenvectors (-1,1,0) and (-1,0,1) ,and for , eigenvector is (1,1,1) . (please check that it is correct).
here is my question, I know that (-1,1,0) and (1,1,1) are orthogonal and so (-1,0,1) and (1,1,1). Also another theorem tells that if and are distinct eigenvalues of T with corresponding eigenvectors x and y , then x and y are orthogonal. but how about (-1,1,0) and (1,1,1)??? Obviously they are not orthogonal. Is it because of that they are eigenvectors of same eigenvalues??
c. here is the definition of the positive definite.
A linear operator T on a finite-dimensional inner product is called positive definite if T is self-adjoint and <T(x),x> >0 for all .
An n x n matrix A with entries from R or C is called positive definite if is positive definite.
Please represent the quadratic form by a symmetric matrix. And verify that Q positive definite.
Orthogonal basis is investigated. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.