First let us convince ourselves that there exist matrices with complex eigenvalues.
Example. Consider the matrix
The trick is to treat the complex eigenvalue as a real one. Meaning we deal with it as a number and do the normal calculations for the eigenvectors. Let us see how it works on the above example.
We will do the calculations for
The associated eigenvectors are given by the linear system
Remark. It is clear that one should expect to have complex entries in the eigenvectors.
We have seen that (1-2i) is also an eigenvalue of the above matrix. Since the entries of the matrix A are real, then one may easily show that if
is a complex eigenvalue, then its conjugate
is also an eigenvalue. Moreover, if X is an eigenvector of A associated to ,
then the vector ,
obtained from X by taking the complex-conjugate of the entries of X, is an eigenvector associated to
So the eigenvectors of the above matrix A associated to the eigenvalue (1-2i) are given by
Let us summarize what we did in the above example.
Summary: Let A be a square matrix. Assume is a complex eigenvalue of A. In order to find the associated eigenvectors, we do the following steps:
In general, it is normal to expect that a square matrix with real entries may still have complex eigenvalues. One may wonder if there exists a class of matrices with only real eigenvalues. This is the case for symmetric matrices. The proof is very technical and will be discussed in another page. But for square matrices of order 2, the proof is quite easy. Let us give it here for the sake of being little complete.
Consider the symmetric square matrix
Remark. Note that the matrix A will have one eigenvalue, i.e. one double root, if and only if
But this is possible only if a=c and b=0. In other words, we have
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Author: M.A. Khamsi