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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.

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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 $A=\color{red}{P^{-1}IP=I}$

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

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  • $\begingroup$ We haven't really done Diagonalization in class, so I'm not sure how it helps. All I understood is that A will be diagonalizable. $\endgroup$
    – Sahiba Arora
    Commented Sep 23, 2015 at 9:22
  • $\begingroup$ So $A$ is diagonalizable, and as you pointed out, $1$ is the only eigenvalue... $\endgroup$
    – Augustin
    Commented Sep 23, 2015 at 9:24
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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 ...

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  • $\begingroup$ I thought if A^2=I, then minimal polynomial divides (x^2-1) and hence we get (x-1) is a factor of minimal polynomial. But how does it show that (x-1) is the minimal polynomial? $\endgroup$
    – Sahiba Arora
    Commented Sep 23, 2015 at 9:58
  • $\begingroup$ @SahibaArora If the minimal polynomial contains $x+1$ as a factor, it would mean $-1$ is an eigenvalue of $A$, but this is impossible because $A$ is positive definite. $\endgroup$
    – user1551
    Commented Sep 23, 2015 at 10:01
  • $\begingroup$ Yes, I get that. But here we know A^2-1=0, can't A satisfy some other equation also, which gives us another eigenvalue? $\endgroup$
    – Sahiba Arora
    Commented Sep 23, 2015 at 10:04
  • $\begingroup$ @SahibaArora No. The minimal polynomial of a matrix $A$ is the unique monic polynomial that divides every annihilating polynomial of $A$. Now $x^2-1$ is an annihilating polynomial of $A$ (because $A^2-I=0$). Since the minimal polynomial must divide $x^2-1$, its factors must be taken from the factors of $x^2-1$. $\endgroup$
    – user1551
    Commented Sep 23, 2015 at 10:13
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    $\begingroup$ Any two annihilating polynomials must share at least one common factor. In your example, $x-2$ and $x^2-1$ have no common factor. So it's impossible that $x-2$ is an annihilating polynomial of $A$. $\endgroup$
    – user1551
    Commented Sep 23, 2015 at 10:25

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