# Suppose $A$ is symmetric matrix. If all the eigenvalues of A are $\pm 1$, show that $A$ is an orthogonal matrix.

Question states: "Suppose $A$ is symmetric matrix. Show that the following are equivalent:

(1) $A$ is orthogonal.

(2) $A^2=I$

(3) All eigenvalues of $A$ are $\pm 1$"

I proved that 1 and 2 are equivalent, and that 1 or 2 implies 3. It remains to prove that 3 implies either 1 or 2.

Here is my proof: since $A$ is symmetric, it is orthogonally diagonalizable. We'll work in the basis such that $A$ diagonal. Then $A$ has all its diagonals as $\pm 1$ and all its nondiagonal entries as $0$, and so $A$ is orthogonal. Does this prove make sense? Am I really safe to assume without loss of generality that $A$ is diagonal? If I am wrong, could you provide an explanation for why 3 implies 1 or 2?

• $A^2=AA=AA'=I\Rightarrow A$ is orthogonal, in other words a symmetric and involutory matrix is orthogonal.(so, $(2)\Rightarrow (1)$). It is a simple exercise now to show the eigenvalues of a real orthogonal matrix are $\pm 1$, which has been shown on this forum before. – StubbornAtom Mar 20 '17 at 4:54

Since $A$ is symmetric, there is an invertible matrix $Q$ such that $Q^{-1}AQ=D$, where $D$ is a diagonal matrix.

And since all the eigenvalues of $A$ are $\pm 1$, it follows that the diagonal entries of $D$ are $\pm 1$, so $D^2=I$. Therefore $$A^2=QD^2Q^{-1}=QQ^{-1}=I$$ so (3) implies (2).