1. ## Nilpotent operator

Let $\displaystyle T:V\to V$ be linear transformation over field, and let $\displaystyle p$ be a polynomial over that field.
Prove that $\displaystyle p(T)$ nilpotent iff $\displaystyle p(0)=0$

Thanks!

2. Originally Posted by Also sprach Zarathustra
Let $\displaystyle T:V\to V$ be linear transformation over field, and let $\displaystyle p$ be a polynomial over that field.
Prove that $\displaystyle p(T)$ nilpotent iff $\displaystyle p(0)=0$

Thanks!
Since p is a polynomial, we can write $\displaystyle p(x) = \sum_{i=0}^n a_ix^i$. Now, note that $\displaystyle p(0) = a_0 , ~ p(0) = 0 \Rightarrow a_0 = 0$. Conclude by using the fact that T is nilpotent.

3. Originally Posted by Defunkt
Since p is a polynomial, we can write $\displaystyle p(x) = \sum_{i=0}^n a_ix^i$. Now, note that $\displaystyle p(0) = a_0 , ~ p(0) = 0 \Rightarrow a_0 = 0$. Conclude by using the fact that T is nilpotent.
Yes, I'm aware this fact.

i) $\displaystyle p(0)=0 \implies p(T)$ nilpotent.
$\displaystyle p(T)=a_1T+a_2T^2+ \dots +T^n$
Above, sum of nilpotent operators $\displaystyle \implies p(T)$ nilpotent.

ii) $\displaystyle p(T) nilpotent \implies p(0)=0$
So, there exist $\displaystyle r$, index of nilpotents so that$\displaystyle {p(T)}^r=0 \implies a_0=0$

4. Originally Posted by Also sprach Zarathustra
Yes, I'm aware this fact.

i) $\displaystyle p(0)=0 \implies p(T)$ nilpotent.
$\displaystyle p(T)=a_1T+a_2T^2+ \dots +T^n$
Above, sum of nilpotent operators $\displaystyle \implies p(T)$ nilpotent.
This is good.

ii) $\displaystyle p(T) nilpotent \implies p(0)=0$
So, there exist $\displaystyle r$, index of nilpotents so that$\displaystyle {p(T)}^r=0 \implies a_0=0$
This needs some justifications (simple), and otherwise you are done.

5. Originally Posted by Also sprach Zarathustra
i) $\displaystyle p(0)=0 \implies p(T)$ nilpotent.
$\displaystyle p(T)=a_1T+a_2T^2+ \dots +T^n$
Above, sum of nilpotent operators $\displaystyle \implies p(T)$ nilpotent.
Maybe you should say "sum of commuting nilpotent operators $\displaystyle \implies p(T)$ nilpotent." It's not true in general that a sum of nilpotent operators is nilpotent.