In another forum (in Spanish : Aplicación diagonalizable - Foro fmat.cl), one asked if , is diagonalizable.
Another guy answered that no, because the geometric multiplicity and the algebraic multiplicity are not equal.
Forming a matrix with T(1,0) and T(0,1) as column-vectors gives eigenvalues of 1. (2 eigenvalues that is).
Then the guy says that it implies that the algebraic multiplicity of 1 is 2. I don't understand the "of 1", and also in another forum I read that the algebraic multiplicity is the multiplicity of the root of the characteristic polynomial, so I understand why here it is 2.
But then the guy says that the geometric multiplicity is 0 because the dimension of the eigenspace is 0. But he never explained why it is 0. I don't see why. Can you help me?
Another question : if I take the identity matrix . Clearly the eigenvalues of it are 1, 1 and 1. So the algebraic multiplicity is 3. But the geometric multiplicity seems to be 1. Hence the identity matrix is not diagonalizable... I don't think it's right. I'm having trouble understanding how do you find the dimension of the eigenspace. Thanks in advance.