Calculating $A^{80}$ for conditions on $A$

Question

Let $A \in \mathbb{C}_{4x4}$ be a matrix fulfills the next conditions -

1) $trace(A) = 0$, 2) $|A-iI| = 0$, 3) $rank(A-I) = 3$, 4) $Av=-v$ for some $v \not= 0.$

Calculate $A^{80}.$

Attempt -

From the 2nd conditions i can say that $i$ is an eigenvalue of $A$, from the 3rd conditions i can conclude that $1$ is an eigenvalue of A, and from the 4th+1st conditions $-1,-i$ are also eigenvalues.

So $A$ is diagonalizable $\implies A = PDP^{-1}$ for some $P$ and $D = diag(1,-1,i,-i)$.

Can i just say that $A^{80} = diag(1, 1, -1 ,1)$, do i even need $P$ and $P^{-1}$ ?

Thank you !

Answer 1

As you mentioned, there are $4$ distinct eigenvalues for this matrix, so that we can find a transformation $P$ with $$ A=PDP^{-1}$$ and $D$ a diagonal matrix, with diagonal entries $1,-1,i,-i$. Then $$A^{n}=\underbrace{(PDP^{-1})\cdots(PDP^{-1})}_{n\:\text{times}}=PD^nP^{-1}.$$ Now, $D^{80}=I_n$, where $I_n$ is the $n\times n$ identity matrix. So, we have that $$ A^{80}=PI_nP^{-1}=I_n.$$ So, $A^{80}=I_n$.

Answer 2

The characteristic polynomial of $A$ is $(x-1)(x+1)(x-i)(x+i)=x^4-1$ and so $A^4=I$. Therefore, $A^{80}=(A^4)^{20}=I$.