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    MHF Contributor arbolis's Avatar
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    Diagonalizable transformation question

    In another forum (in Spanish : Aplicación diagonalizable - Foro fmat.cl), one asked if T:\mathbb{R}^2 \to \mathbb{R}^2 , 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 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.
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    Quote Originally Posted by arbolis View Post

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

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

    obviously generated by (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.
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    MHF Contributor arbolis's Avatar
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    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 \{ (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 T(x,y,z)=(x,y,z). If it is right then the basis I just gave is right and I understand everything.
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    Quote Originally Posted by arbolis View Post
    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 \{ (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 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 A is given, the corresponding transformation is defined by: Tv=Av. so if A=I, the identity matrix, then Tv=Iv=v, which is what you guessed.
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