eigenvalue , auch: characteristic root , auch: characteristic value

So and share the same eigenvalues by the discussion above.

So far, we have restricted our studies to real eigenvalues.

The equation is true even when the eigenvalues are not all different.

Care must be taken since some of the eigenvalues could have the same magnitude.

Actually the number of such eigenvalues is exactly equal to the period.

Once again we here assume the eigenvalues s to be all different.

This is the case when the eigenvalues have the same magnitude.

This process can be repeated until all eigenvalues are found.

Thus any projection has 0 and 1 for its eigenvalues.

Only 0 and 1 can be an eigenvalue of a projection.

This is a contradiction, and so A has an eigenvalue.

If it is negative then the two eigenvalues have different signs.

It no longer makes sense here to list the positive eigenvalues in increasing order.

The eigenvalues of this matrix are therefore 20 and 25.

This is seen most easily by looking at the problem of zero eigenvalues.

The first step, that has already been mentioned above, is finding the eigenvalues.

Both of these two circumstances result in an eigenvalue equal to one.

A natural way to do this is by eigenvalue analysis of a matrix.

We'll use the calculated eigenvalues later in the final solution.

This example covers only the case for real, separate eigenvalues.

However, the spin can take only two different values (eigenvalues).

The number 0 is not an eigenvalue of A.

This is not physically a problem, the real state might have zero eigenvalues.

Curiously, the situation is not at all bad if two or more eigenvalues are exactly equal.

R is minimum when the eigenvalues are equal to each other.