1. ## Diagonalizable transformation question

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 explained why 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.

2. Originally Posted by arbolis

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?
that is not correct! the dimesnion of the eigenspace in this example is 1 not 0. the dimesnion of the eigenspace corresponding to an eigenvalue is never 0. it's always at least 1.

the eigenspace corresponding to the eigenvalue $\displaystyle \lambda$ of a transformation $\displaystyle T$ is the solution set of $\displaystyle Tv=\lambda v.$ for example, in the above example, you need to solve $\displaystyle T(x,y)=(x,y),$

because $\displaystyle \lambda=1.$ so we'll have $\displaystyle (x+y,y)=(x,y),$ which gives us $\displaystyle y=0$ and $\displaystyle x$ can be anything. so the eigenspace corresponding to $\displaystyle \lambda=1$ is the set $\displaystyle \{(x,0): \ x \in \mathbb{R} \},$ which is

obviously generated by $\displaystyle (1,0).$ so {(1,0)} is a basis for the eigenspace and thus the dimension of the eigenspace is 1. i think you can now answer your second question yourself.

3. Thanks, that was very clear.
However I'm stuck to answer my second question. My intuition says that the geometric multiplicity is 3 because a basis for the eigenspace is $\displaystyle \{ (1,0,0), (0,1,0), (0,0,1) \}$. I'm stuck to see what is the transformation associated to the identity matrix. I guess it's something like $\displaystyle T(x,y,z)=(x,y,z)$. If it is right then the basis I just gave is right and I understand everything.

4. Originally Posted by arbolis
Thanks, that was very clear.
However I'm stuck to answer my second question. My intuition says that the geometric multiplicity is 3 because a basis for the eigenspace is $\displaystyle \{ (1,0,0), (0,1,0), (0,0,1) \}$. I'm stuck to see what is the transformation associated to the identity matrix. I guess it's something like $\displaystyle T(x,y,z)=(x,y,z)$. If it is right then the basis I just gave is right and I understand everything.
correct! in general, if a matrix $\displaystyle A$ is given, the corresponding transformation is defined by: $\displaystyle Tv=Av.$ so if $\displaystyle A=I,$ the identity matrix, then $\displaystyle Tv=Iv=v,$ which is what you guessed.