In another forum (in Spanish : Aplicación diagonalizable - Foro fmat.cl), one asked if $\displaystyle T:\mathbb{R}^2 \to \mathbb{R}^2$ , $\displaystyle T(x,y)=(x+y,y)$ 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 explainedwhy it is 0. I don't see why. Can you help me?

Another question : if I take the identity matrix $\displaystyle 3 \times 3$. 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.