1. Vector Spaces

Let V be a complex Vector Space, and a:V->V be a linear map.

a) Show that is a has finite order, then it is diagonalisable.
b) Let v in V be non-zero. Show that if there exists a smallest integer s S.T. a^s(v) = 0, then v, a(v), a^2(v),..., a^s-1(v) are Lin. Independent.

I'm really struggling with this one on a problem sheet here. Can anyone help?

2. Originally Posted by bumcheekcity
Let V be a complex Vector Space, and a:V->V be a linear map.

a) Show that is a has finite order, then it is diagonalisable.
In the Jordan normal form of A, the diagonal entries will all be roots of unity, and the off-diagonal elements must all be zero (because when you take powers of it, the off-diagonal elements can only get bigger). So A has a diagonal JNF and is therefore diagonalisable.

Originally Posted by bumcheekcity
b) Let v in V be non-zero. Show that if there exists a smallest integer s S.T. a^s(v) = 0, then v, a(v), a^2(v),..., a^s-1(v) are Lin. Independent.
Suppose that $\displaystyle \lambda_1Av + \lambda_2A^2v + \ldots + \lambda_{s-1}A^{s-1}v = 0$. Multiply both sides by $\displaystyle A^{s-2}$ and you see that $\displaystyle \lambda_1A^{s-1}v = 0$, hence $\displaystyle \lambda_1 = 0$.

Therefore $\displaystyle \lambda_2A^2v + \ldots + \lambda_{s-1}A^{s-1}v = 0$. Multiply both sides by $\displaystyle A^{s-3}$ and you see that $\displaystyle \lambda_2 = 0$. And so on. So all the coefficients are zero and thus the vectors are lin. ind.

3. Thanks!