Show that if A is an n-by-n real symmetric matrix with $A^k$=Id for some k$\ge$1. Then $A^2$=Id. My attempt
Proof. We know that if $A^k$=id for some k$\ge$$1$, then A=\begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} or $A^k$=\begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} for some k$>$$1$ by properties of matrix multiplication thus for k=2, we have that $A^2$=(id)(id)=(id), as claimed.
Note that id denotes the identity matrix.
I'm not sure whether my approach here is correct, but I argued that given the problem statement whereby $A^k$=id for some k at least 1, this means that k is at least 1 s.t. $A^1$=id holds? Hence if $A^1$=id, then by properties of matrix multiplication $A^2$=id. I may be interpreting this wrong, but my understanding is that the "OR" gate has 75% of the sample space available. In other words, if some k=1 is true and some k>1 is true, then using the fact that $A^1$=id, can we reason that $A^2$=id?
 A: Edit: The source of your confusion seems to stem from the phrase "for some $k\geq1.$" You are correctly understanding the $k\geq 1$ part--this means $k$ could be $1,2,3,4,5,$ etc.--but it seems that you're thinking we're allowed to choose $k=1.$ This is absolutely false. For a simple example, consider $A=-I,$ so that $A^1\neq I,$ but $A^2=I.$
Rather, we must consider $k$ to be a completely arbitrary positive integer. To put it another way, we could consider this claim to be the following collection of related claims:

*

*If $A$ is an $n\times n$ real symmetric matrix and $A^1=I,$ then $A^2=I.$


*If $A$ is an $n\times n$ real symmetric matrix and $A^2=I,$ then $A^2=I.$


*If $A$ is an $n\times n$ real symmetric matrix and $A^3=I,$ then $A^2=I.$


*If $A$ is an $n\times n$ real symmetric matrix and $A^4=I,$ then $A^2=I.$


*If $A$ is an $n\times n$ real symmetric matrix and $A^5=I,$ then $A^2=I.$


*If $A$ is an $n\times n$ real symmetric matrix and $A^6=I,$ then $A^2=I.$
and so on.

Added: Let me give some terminology that might clear things up.
Given a square matrix $B,$ if there is no integer $k\ge 1$ for which $B^k=I,$ we say that $B$ has infinite order. Otherwise, we say that $B$ has finite order, and define the order of $B$ to be the smallest integer $k\ge 1$ for which $A^k=I.$
The claim states that if a real symmetric matrix $A$ has finite order, then it either has order $1$ (meaning that $A=I$) or has order $2$ (meaning that $A\neq I$ and $A^2=I$).
As a consequence of this, if I were to tell you I was thinking of a non-identity real symmetric matrix $A,$ you would immediately be able to conclude that $A^k\neq I$ whenever $k$ is a positive odd integer, despite knowing literally nothing else about my matrix. If I told you further that I calculated $A^{2640887552}$ and found out that it was the identity matrix, then (assuming I wasn't lying) you could conclude that $A^2=I,$ despite not knowing what my matrix is, or even what size it is! Moreover, you could conclude that $A^k=I$ for all positive even integers $k.$
