<s>[INST] Which of the following statements are false Let$$A=\frac{1}{3}\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 &  2  \end{pmatrix}$$
Which of the following statements are false?
(a) $A$ has only one real eigenvalue.
(b) $\operatorname{Rank}(A) = \operatorname{Trace}(A)$.
(c) Determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$.
I proceed in this way:
(a) let $\lambda$ be the eigenvalue of $A$. So,$[A-\lambda I]=\begin{pmatrix} 2/3-\lambda & -1/3 & -1/3 \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 &  2/3-\lambda  \end{pmatrix}=\begin{pmatrix} -\lambda & -\lambda & -\lambda \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 &  2/3-\lambda  \end{pmatrix}$$[R_1'\to R_1+R_2+R_3]$$$=\begin{pmatrix} -\lambda & 0 & 0 \\ -1/3 & 1-\lambda & 0 \\ -1/3 & 0 &  1-\lambda  \end{pmatrix}[C_2'\to C_2-C_1, C_3'\to C_3-C_1]$$ Hence we get $$|A-\lambda I|=0$$$=>\lambda(1-\lambda)^2=0$ $=>\lambda= $ $0$ or $1$ $=>$There are two real eigenvalue.   
(b) Note that $|A|=0$ and $$\begin{vmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{vmatrix}=5/9 \not= 0$$ So , $\operatorname{Rank}(A) =2$ and $\operatorname{Trace}(A)=\frac{1}{3}(2+2+2)=2$ Hence $\operatorname{Rank}(A) = \operatorname{Trace}(A)$   
(c) $A^2=AA=A$ and $A^3=A^2A=AA=A$ so determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$. So only first statement is wrong.
 [/INST] You are totally right, except the third part. You do not have $A^2=A$, but instead you have that $det(A^n)=(det(A))^n=0^n=0=det(A)$.
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