4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Sep 23 '15 at 9:22
  • $\begingroup$ So $A$ is diagonalizable, and as you pointed out, $1$ is the only eigenvalue... $\endgroup$
    – Augustin
    Sep 23 '15 at 9:24
1
$\begingroup$

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 ...

$\endgroup$
9
  • $\begingroup$ 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? $\endgroup$ Sep 23 '15 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
    Sep 23 '15 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? $\endgroup$ Sep 23 '15 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
    Sep 23 '15 at 10:13
  • 1
    $\begingroup$ 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
    Sep 23 '15 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.