Let $ V$ be a vector space over $\mathbb{C}$ with dimension $n$. Then which of the following must be true?

Question

let $ V$ be a vector space over $\mathbb{C}$ with dimension n. Let $T :V \rightarrow V $ be a linear transformation with only $1$ as eigenvalue .then which of the following must be TRue ?

$1)$. $T - I=0$

$2)$. $(T-I)^{n-1} =0$

$3)$.$(T-I)^n=0$

$4)$.$(T -I)^{2n} =0$

My attempts : i take $T = I_{m \times n}$ where I is Identity matrix

then option $1)$ and option $3)$ is Correct

now see option $2)$ is not coorect because put $n =1$ we get $(T-I)^{1-1}=0$, $(T-I)^{0}= 1\neq 0$

As option $4$ is also not correct as Put $n=1$ as $(T-I)^{2} =(T-I) (T-I)=0$, now we claim that either$(T-I)=0 $ or ($T-I)=0$ boths can not be $ 0$

Is my answer is correct or not ?

Pliz tell me..

Any hints/solution will be appreciated.

thanks u

Answer 1

No, it is not correct. Yes, assertion 1) holds for the identity, but it is not always true. For instance, if $f(z,w)=(z+w,w)$, then $1$ is the only eigenvaiue of $f$, but 1) doesn't hold.

Actually, 3) and 4) always hold. Since $1$ is the only eigenvaly, the characteristic polynomial of $T$ is $(1-\lambda)^n$. It follows from the Cayley-Hamilaton that $(T-\operatorname{Id})^n=0$. And $$(T-\operatorname{Id})^n=0\implies(T-\operatorname{Id})^{2n}=\left((T-\operatorname{Id})^n\right)^2=0.$$