Identity is the only real matrix which is orthogonal, symmetric and positive definite

Question

Show that identity is the only real matrix which is orthogonal, symmetric and positive definite

All I could get using above information was that $A^2=I$, hence it is its own inverse.

Using the fact that $A$ is positive-definite, I got that all diagonal entries will be greater than $0$, but how does that help?

Edit: As $A$ satisfies $(x^2-1)=0$, therefore the minimal polynomial will divide this. Therefore, the minimal polynomial will have $(x-1)$ or $(x+1)$ or both as a factor, as $A$ is positive definite, so $-1$ can't be an eigenvalue, therefore we get that $1$ is an eigenvalue of $A.$

I'm not sure, if this helps, though.

Answer 1

there is unitary matrix $P$ and diagonal matrix $D$ such that $\color{red}{A=P^{-1}DP}$ thus combine it with $A^2=I$ you have $I=A^2=P^{-1} D P P^{-1}DP=P^{-1}D^2 P$ hence $D^2=PIP^{-1}=I$ thus entry of matrix $D$ are as $a^2=1\Rightarrow‎ a=\pm 1$ which since $A$ is positive definite only $a=1$ is acceptable. this means $D=I$ thus $A=\color{red}{P^{-1}IP=I}$

Answer 2

Hint: $A$ is symetric and positive-definite. Use diagonalization.

Answer 3

An alternative approach: as $A^2=A^TA=I$, the minimal polynomial of $A$ is either $x-1,\ x+1$ or $x^2-1$. Yet all eigenvalues of $A$ are positive (because $A$ is positive definite). So, the minimal polynomial must be ...