$3\times 3$ real orthogonal matrix $A$ with characteristic polynomial $(x+1)(x-1)^2$ is symmetric 
Suppose that $A$ is a $3\times 3$ real orthogonal matrix and the characteristic polynomial of $A$ is $(x+1)(x-1)^2$ . Prove that $A$ is symmetric.


I know that $A$ is a real normal matrix with real eigenvalues and hence symmetric, see A normal matrix with real eigenvalues is Hermitian
.
But I think that's overkill. Since we did not even use the conditions on eigenvalues. So is there an elementary method to tackle this question? Thank you.
 A: Note that $A^2$ is real and orthogonal too, with $1$ its only eigenvalue. The only such matrix is $I$. Therefore,
$$A^T = A^T I = A^T AA = IA = A.$$
A: Here is a proof without using any spectral property of orthogonal or normal matrices, but it only applies because $A$ is $3\times3$.
By assumption, $1$ and $-1$ are eigenvalues of $A$. Therefore there exist real unit vectors $x$ and $y$ such that $Ax=x$ and $Ay=-y$. Then $y^Tx=y^TIx=y^TA^TAx=-y^Tx$. Hence $y^Tx=0$, i.e. $x\perp y$. Complete $\pmatrix{x&y}$ to a real orthogonal matrix $Q=\pmatrix{x&y&z}$. Since $A$ and $Q$ are real orthogonal, so is $AQ=\pmatrix{x&-y&Az}$. Hence $Az$ is orthogonal to both $x$ and $y$ and it must be a scalar multiple of $z$. In other words, $z$ is also an eigenvector of $A$. Thus $A$ has an orthonormal eigenbasis over $\mathbb R$, meaning that $A$ is real symmetric.
A: Since $A$ is orthogonal and has real eigenvalues hence must be diagonalizable which implies that minim polynomial must be $m(t)=(t-1)(t+1)=t^2-1$.
So $A^2-I=0\implies A^2=I\implies AA=I$.....(1)
By orthogonality of $A$, we already have $AA'=I$.....(2)
From (1) and (2), $A=A'$ follows.
