# Diagonalizable transformation question

• Jun 26th 2009, 05:32 PM
arbolis
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.
• Jun 26th 2009, 05:55 PM
NonCommAlg
Quote:

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.
• Jun 26th 2009, 06:10 PM
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.
• Jun 26th 2009, 06:21 PM
NonCommAlg
Quote:

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.