I ve been working on this exercise unsuccessfully for some time. I hope my translation to english is clear enaugh for you to help me.
I ve solved items 1 ( -1,0, 1 are the eigenvalues ) and 2b that is trivial, but none of the rest.
Here it goes:
Let V be a (real) vector space with finite dimension and T an operator $T:V \mapsto V $ such that $T=T^{3}$ that has three eigenvalues $\lambda _1 < \lambda _2 < \lambda _3 $ with $S_{\lambda_1},S_{\lambda_2},S_{\lambda_3}$ the corresponding eigenspaces.
- Find $\lambda _1 , \lambda _2 , \lambda _3 $ With $v \in V$
- Prove that a) $\frac{1}{2}T^{2}(v) - \frac{1}{2}T(v) \in S_{\lambda_1}$ b) $v - \frac{1}{2}T^{2}(v) \in S_{\lambda_2}$ c) $\frac{1}{2}T^{2}(v) + \frac{1}{2}T(v) \in S_{\lambda_3}$
- Prove that $V= S_{\lambda_1}\oplus S_{\lambda_2}\oplus S_{\lambda_3}$
- Prove that T is diagonalizable
- Find the possible diagonal matrixes associated with T if $dim(V)=4$
Thanks in advance
Edit 1 :
Thanks all for the tips I ve solved parts 1,2,3,4 (I searched for the decomposition theorem but didn t use it )
For proving the second part just did the calculations.
For proving the direct sum, I used that for any v in V, adding the a), b) c) statements , every vector in V is in S1+S2+S3, as S1,S2,S3 just have the zero element in common, it has to be a direct sum.
For proving that it should then be diagonalizable operator , i used that $dim (V) = dim (S1) + dim (S2) + dim (S3)$ So the associated eigenvalues have the algebraic multiplicity needed to split the characteristic polynomial , and then there is a diagonal matrix associated to T.
For the last fifth part I still have some doubts....
¿Can I say, having in $ker(T)$ vectors such as $v=T^{2}(v)$ , (as proved in prior part), that $Kert(T)$ has then dimension 2 at least, because $T^{2}$ "inherits" the eigenvalues 1 and 0 from T?
If so, all the possible matrices with $dim(V)= 4 $are just those having the elements 1,-1,0,0 in the main diagonal .
That makes 12 possible 4×4 matrices, if I counted them right.