Positive definiteness of an inverse

I would appreciate it if someone could just verify whether this works or not.

Prove that if T is positive definite, then T^{-1} is positive definite.

Since $\displaystyle \langle T(x),x \rangle > 0$ and $\displaystyle T(T^{-1}(x))=x$, we have $\displaystyle 0 < \langle T(x),x \rangle = \langle T(x),T(T^{-1}(x)) \rangle = \langle T^{-1}(T(x)),T^{-1}(T(T^{-1}(x))) \rangle$

$\displaystyle = \langle x,T^{-1}(x) \rangle = \langle T^{-1}(x),x \rangle$. The last equality is true since $\displaystyle \langle T(x),x \rangle$ is real, and also proves that $\displaystyle T^{-1}$ is self-adjoint, and thus is positive definite.

Thanks.

Re: Positive definiteness of an inverse

Quote:

Originally Posted by

**AlexP** I would appreciate it if someone could just verify whether this works or not.

Prove that if T is positive definite, then T^{-1} is positive definite.

Since $\displaystyle \langle T(x),x \rangle > 0$ and $\displaystyle T(T^{-1}(x))=x$, we have $\displaystyle 0 < \langle T(x),x \rangle = \langle T(x),T(T^{-1}(x)) \rangle = \langle T^{-1}(T(x)),T^{-1}(T(T^{-1}(x))) \rangle$

$\displaystyle = \langle x,T^{-1}(x) \rangle = \langle T^{-1}(x),x \rangle$. The last equality is true since $\displaystyle \langle T(x),x \rangle$ is real, and also proves that $\displaystyle T^{-1}$ is self-adjoint, and thus is positive definite.

Thanks.

This looks fine, but it may have been easier to just note that being positive definite is equivalent to having strictly positive eigenvalues, and since the eigenvalues of $\displaystyle T^{-1}$ are the reciprocals of the eigenvalues of $\displaystyle T$ everything quickly follows.

Re: Positive definiteness of an inverse

[QUOTE=AlexP;705251]$\displaystyle \langle T(x),T(T^{-1}(x)) \rangle = \langle T^{-1}(T(x)),T^{-1}(T(T^{-1}(x))) \rangle$

I'm afraid this is not generally true.

It requires $\displaystyle T^{-1}$ to preserve the inner product, which is not automatically the case.