Symmetric matrices have some remarkable properties that can be summarized by:
- (a)
- every eigenvalue of is real, and
- (b)
- there is an orthonormal basis of consisting of eigenvectors of .
As a consequence of Theorem ??, let be an orthonormal basis for consisting of eigenvectors of . Indeed, suppose where . Note that It follows from (??) that is a diagonal matrix. So every symmetric matrix is similar to a diagonal matrix.
Hermitian Inner Products
The proof of Theorem ?? uses the Hermitian inner product — a generalization of dot product to complex vectors. Let be two complex -vectors. Define Note that the coordinates of the second vector enter this formula with a complex conjugate. However, if and are real vectors, then A more compact notation for the Hermitian inner product is given by matrix multiplication. Suppose that and are column -vectors. Then
The properties of the Hermitian inner product are similar to those of dot product. We note three. Let be a complex scalar. Then
Let be a complex matrix. Then the main observation concerning Hermitian inner products that we shall use is: This fact is verified by calculating So if is a real symmetric matrix, then
since .
- Theorem ??(a)
- Let be an eigenvalue of and let be the associated eigenvector. Since we can use (??) to compute Since , it follows that and is real.
- Theorem ??(b)
- Let be a real symmetric matrix. We want to show that
there is an orthonormal basis of consisting of eigenvectors of . The proof
proceeds inductively on . The theorem is trivially valid for ; so we assume that
it is valid for .
Theorem ?? of Chapter ?? implies that has an eigenvalue and Theorem ??(a) states that this eigenvalue is real. Let be a unit length eigenvector corresponding to the eigenvalue . Extend to an orthonormal basis of and let be the matrix whose columns are the vectors in this orthonormal basis. Orthonormality and direct multiplication implies that
Therefore is invertible; indeed .Next, let By direct computation It follows that that has the form where is an matrix. Since , it follows that is a symmetric matrix; to verify this point compute It follows that where is a symmetric matrix. By induction we can choose an orthonormal basis in consisting of eigenvectors of . It follows that is an orthonormal basis for consisting of eigenvectors of .
Finally, let for . Since , it follows that is a basis of consisting of eigenvectors of . We need only show that the form an orthonormal basis of . This is done using (??). For notational convenience let and compute
since . Thus the vectors form an orthonormal basis since the vectors form an orthonormal basis.
Exercises
- Prove directly using the discriminant of the characteristic polynomial that has real eigenvalues.
- Show that has equal eigenvalues only if is a scalar multiple of .
In Exercises ?? – ?? compute the eigenvalues and the eigenvectors of the matrix. Then load the matrix into the program map in MATLAB and iterate. That is, choose an initial vector and use map to compute , , …. How does the result of iteration compare with the eigenvectors and eigenvalues that you have found? Hint: You may find it convenient to use the feature Rescale in the MAP Options. Then the norm of the vectors is rescaled to after each use of the command Map and the vectors will not escape from the viewing screen.