I have some trouble with this question, I tried to solve it but I'm not sure that my solution is correct. I'll be glad if somebody could take a look.
Data :
T : R^4 --> R^4 (linear transformation)
Characteristic polynomial --> p(x) = x^4-x^3, and also data that dim(ImT)=2. The question is if T diagonalizable?
That what I did :
Since dim(ImT)=2, and T:R^4 --> R^4 we know that dim(KerT) = 2. As well :
x^4-x^3 --> x^3(x-1)
So 0,1 are eigenvalue.
Since 0,1 are eigenvalue we can conclude that matrix is similar same characteristic polynomial.
So it possible to conclude that:
In case 1 eigenvalue, algebra multiplexing = 3, geometric multiplexing = (4-3) = 1
In case 0 eigenvalue, algebra multiplexing = 1, geometric multiplexing = (4-1) = 3
so T is not diagonalizable.
Thank you!
Your general idea is correct but you made a couple mistakes. $0$ has an algebraic multiplicity of $3$ (and $1$ has an algebraic multiplicity of $1$, though this doesn't matter in this problem.) On the other hand, $0$ has geometric multiplicity of $2$: the eigenspace corresponding to the eigenvalue $0$ is $\text{Ker } T$, which you correctly determined has dimension $2$. Since $2<3$, $T$ is not diagonalizable.
By the way, in general, the geometric multiplicity of an eigenvalue is less than or equal to its algebraic multiplicity. So it is not possible for an eigenvalue to have algebraic multiplicity $1$ and geometric multiplicity $3$.