Now I got the right matrix and it works with the , thank you very much.
There is only one thing that i cant seem to figure out, when i calculate
Using the rule of Sarrus I don't get .
Could you show me how you came to the equation?
/Maria
Ok. In order to compute the eigenvalues, you set the following determinant,
equal to zero. The equation you get is
The power of each of these factors is its algebraic multiplicity. Hence, is raised to the power 1, and therefore, by definition, the eigenvalue has algebraic multiplicity 1. The factor is raised to the power 2, and therefore, by definition, the eigenvalue has algebraic multiplicity 2.
The geometric multiplicity of an eigenvalue is the dimension of the corresponding eigenspace. When you computed the eigenvectors for you found two linearly independent eigenvectors. Hence, the geometric multiplicity of the eigenvalue is 2. I have a very strong feeling that when you find the eigenvector (which you should do without delay!) corresponding to you will find only one eigenvector. Hence, the geometric multiplicity of the eigenvalue is 1.
These are definitions.
Now, I really think you have enough information to solve this problem completely. Keep your head on the target, which is orthogonally diagonalizing
The "algebraic multiplicity" of an eigenvalue is its multiplicity as a root of the characteristic equation. Here, "1" is a double root so it has multiplicity 2. The "geometric multiplicity" is the dimension of the subspace of all eigenvectors with that eigenvalue. Saying that "1" has geometric multiplicity 2 is saying that there the subspace of all eigenvectors with eigenvalue "1" has dimension 2. In particular, that means we could find two independent vectors in that subspace. If there is, as Ackbeet is saying (I have not yet checked it myself), there is another eigenvalue, then we can find a third vector, independent of the other two (eigenvectors corresponding to distinct eigenvalues are always independent). If there are three distinct eigenvectors for a linear tranfromation from to [tex]R^3[/mat], then we can use them as a basis for the vector space and the matrix representing the linear transformation, written in that basis, is diagonal.