is there a non unit real matrix satisfied $A^n=I$?

Question

If A is a real matrix and $A^{2016}$ is a symmetric positive definite matrix , prove that $A$ also is a symmetric positive definite matrix

I wonder if this property is wrong and so I came up with this question : Is there a non unit real matrix satisfied $A^n=I$ for every $n>1$ ?

Answer 1

That's not true.

Consider the matrix $A=\begin{bmatrix} 0&1\\ -1&0 \end{bmatrix}$, where its eigenvalues are $i,-i$. We have that: $$A^{2016} = \begin{bmatrix} 1&0\\0&1 \end{bmatrix},$$ which is positive definite, but $A$ is not even a symmetric matrix.

Answer 2

Original:

is there a non unit real matrix satisfied A^n=I for every n>1 ?

Interpretation:

For each natural number $n>1$, is there always matrix over the real numbers satisfying $A^n=I$ besides the identity matrix?

For any $n$, you can find a primitive $n$'th root of unity in $\Bbb C$ and then represent it as a matrix in $M_2(\mathbb R)$.

You can do this for larger matrices by just embedding the $2\times 2$ in the upper left hand corner.

Answer 3

My interpretation: Does there exist a matrix $A \neq I$ such that for all $n > 1,$ $A^n = I.$ The answer is no, let $A \neq I$ and $A^2 = I.$ Then if $ A(A^2) = A = I,$ we reach a contradiction. Use induction to prove the general result.