Given $(A-I)^2 = 0$, can we say det$(A)=1$ and tr$(A)=n$?

Question

Given $(A-I)^2 = 0$, can we say det$(A)=1$ and tr$(A)=n$?

First I showed when $A$ is two by two, this is true. Since $\lambda_1 + \lambda_2 = 2$ and $\lambda_1 \lambda_2=1$, then from substitution $\lambda_1 = \frac{1}{\lambda_2}$, this gives $(\lambda_2-1)^2 = 1$, and therefore $\lambda_1 = \lambda_2 = 1$.

I tried the case when $A$ is $3\times 3$ by first fixing the third eigenvalue and solve for the other two, and it also worked. How would I approach such problem when $A$ is $n\times n$.

Answer 1

Let $\lambda$ be an eigenvalue (in the complex numbers) of $A$; then $Av=\lambda v$ for some $v\ne0$ in $\mathbb{C}^n$; therefore $$ (A-I)v=(\lambda-1)v $$ and $$ 0=(A-I)^2v=(A-I)(\lambda-1)v=(\lambda-1)^2v $$ Therefore $(\lambda-1)^2=0$ and so $\lambda=1$. Thus $A$ has just the eigenvalue $1$, with algebraic multiplicity $n$.

The determinant of $A$ is the product of the eigenvalues and the trace their sum (counted with their multiplicity).

Answer 2

$(t-1)^2=0$ is anihilating polynomial which suggests two possibilities of the minimal polynomial i.e. $(t-1)=0$ or $(t-1)^2=0$. In either case, the only eigenvalue is $1$. Now, trace(A)$=\sum_{i=0}^{n}t_i$