$Tr(A^2)=Tr(A^3)=Tr(A^4)$ then find $Tr(A)$ Let $A$ be a non singular $n\times n$ matrix  with all eigenvalues real and 
$$Tr(A^2)=Tr(A^3)=Tr(A^4).$$Find $Tr(A)$.
I considered $2\times 2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and tried computing traces of $A^2,A^3,A^4$ and ended up with following 


*

*$Tr(A^2)=Tr(A)^2-2\det(A)$

*$Tr(A^3)=Tr(A)^3-3Tr(A)\det(A)$

*$Tr(A^4)=Tr(A)^4-4Tr(A)^2\det(A)+2\det(A)$


I have no idea how to proceed from here...
 A: A slightly different approach: we have $\operatorname{trace}((A-A^2)^2)=0$. As $A$ has a real spectrum, so does $A-A^2$. So, the previous trace condition implies that $A-A^2$ is nilpotent. Hence every eigenvalue of $A$ is equal to its square. As $A$ is nonsingular, every eigenvalue of $A$ must be equal to $1$. Hence the trace of $A$ is $n$.
A: Let's denote the eigenvalues by $t_j$. Then, by Cauchy-Schwarz
$$
\sum t_j^3 \le \left( \sum t_j^2 \sum t_j^4\right)^{1/2} ,
$$
which equals $\sum t_j^3$ by assumption. Equality in the CSI means that the vectors are linearly dependent, so $t_j=ct_j^2$. This says that there is only one eigenvalue ($=1/c$), and clearly it must be $1$ then. So $\textrm{tr}\, A=n$.
A: we know that trace of a matrix is simply equal to the sum of all its eigen values.
Now, let all the eigen values of the matrix A of order n be as  a_1,a_2,a_3,.....,a_n
Also,we have an important result that for any square matrix A of order n , if  a is an eigen value of A then a^m is an eigen value of the matrix A^m ,for any positive integer m
Applying the above result,we get the following:
(a_1)^2,(a_2)^2,(a_3)^2,..........(a_n)^2 is the entire list of eigen values of the matrix  A^2
likewise,(a_1)^3,(a_2)^3,(a_3)^3,..........(a_n)^3 is the entire list of eigen values of the matrix  A^3
(a_1)^4,(a_2)^4,(a_3)^4,..........(a_n)^4 is the entire list of eigen values of the matrix  A^4
Now,from the given condition in the question,we have
(a_1)^2+(a_2)^2+(a_3)^2,..........+(a_n)^2 = (a_1)^3+(a_2)^3+(a_3)^3+..........+(a_n)^3 = (a_1)^4+(a_2)^4+(a_3)^3+..........+(a_n)^4      _______________(***)
Now, a_1,a_2,......,a_n are all real numbers(mentioned in pblm)
using the condition (***), we can easily deduce the relation 
{a_1(a_1-1)}^2 + {a_2(a_2-1)}^2 + {a_3(a_3-1)}^2 +..........+{a_n(a_n-1)}^2 =0-_________________($$$)
which implies that each of the terms on L.H.S of the above equation equals 0 ,because each of the terms on L.H.S of the above eqn is a real no. and the sum of two or more real squares equals 0 iff each of them is individually 0
But,none of the eigen values of the matrix A is 0 because the det value of a matrix being equal to its product of eigen values would then become 0,contradiction to the given fact that A is non-singular
Hence,from equation($$$), we get that each eigen value of the matrix A is equal to 1.
So,trace of A = sum of all its eigen values = n (answer:)
