Given T:R^n -> R^n , T(vn)=v1 and for $ n-1\ge i \ge 1$ $T(v_i)=v_{i+1}$ T is diagonalizable ? if yes ,than show [T]E Diagonal

Question

Given $$B=\left \{v_1,v_2,...,v_n \right \}$$ a Base in $R^n$

Given$$ T:R^n -> R^n$$ , T(vn)=v1 and for $ n-1\ge i \ge 1$ $T(v_i)=v_{i+1}$ T is diagonalizable ? if yes ,than show [T]E where E is the standard Basis for $R^n$

i feel i dont get to the buttom of the answer

my try was : when$ n=k , k>3$ than : $p_A(t) = t^n+1 $ and so the eigen values is : i,-i,i...,i

than $D=diag(i,-i,i,-i,...,i)$ T is diagonalizable over $C $ \

when for n=3 $$t=1, \frac{-1\pm\sqrt{3}i}{2}$$

then

$D2=diag(1,z,\bar{z},1,z,\bar{z},...,1) $

am i correct ? can some one give a full proof on this question ?

General i assume will be : $$\det(A-tI)=(-1)^n\left(t^n\displaystyle-1^{n}\right)n>1,$$

Edit: when $n>2$

Answer 1

Assume $B=\{v_{1},...,v_{n}\}$ basis for $\mathbb{R}^n$. Then, following $T(v_{k})$ for each $1\leq k \leq n$, we then get:

$[T]_{B}=\begin{pmatrix} 0 &0&0&...& 1\\1&0&0&...&0\\.&&&&.\\.&&&&.\\.&&&&.\\0&0&...&1&0 \end{pmatrix}$

Now if you prove $[T]_{B}$ is diagnolizable then you prove T is. Can you take it from here?

Answer 2

Put $v_{n+1}=v_1$. Let $\zeta$ be any of $n$ complex roots of $1$ and $v=\sum_{k=1}^n \zeta^{-k} v_k$. Then $$Tv=\sum_{k=1}^n \zeta^{-k} v_{k+1}=\zeta\sum_{k=1}^n \zeta^{-k-1}v_{k+1}=\zeta v.$$

So $\zeta$ is an eigenvalue of $T$. Since $T$ has $n$ distinct eigenvalues, its matrix is diagonalizable over $\Bbb C$.