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Let $\alpha$ be a positive definite automorphism of an inner product space $V$. Is $\alpha^{-1}$ necessarily positive definite?

I know the answer is true for invertible positive definite matrices, but I tried to prove this problem for the general case. However, still, I feel that I'm missing something.

Since $\alpha$ is positive definite, it's self-adjoint and for all nonzero $v\in V$, we have $\langle\alpha(v),v\rangle> 0.$

Let $v\in V$, so $$\langle \alpha^{-1}(v),v\rangle=\langle \alpha^{-1}(v),\alpha\alpha^{-1}(v)\rangle=\langle \alpha\alpha^{-1}(v),\alpha^{-1}(v) \rangle > 0$$

Now for each $v\in V$, we got $\langle \alpha^{-1}(v),v\rangle>0$, but still I need to show $\alpha^{-1}$ is self-adjoint. For finitely generated inner space, I can show that it's true.

Let $v, \:w\in V$. Then,

$$\langle v,w\rangle=\langle \alpha^{-1}\alpha(v),w\rangle=\langle\alpha(v), (\alpha^{-1})^*(v)\rangle=\langle v,\alpha^*(\alpha^{-1})^*(v)\rangle, $$ so $(\alpha^{-1})^*=(\alpha^*)^{-1}. $ Moreover, $$\langle\alpha^{-1}(v),w\rangle=\langle v,(\alpha^{-1})^*(w)\rangle=\langle v,(\alpha^*)^{-1}(w)\rangle=\langle v,\alpha^{-1}(w)\rangle.$$ Hence $\alpha^{-1}$ is self-adjoint.

But how about when $V$ is not finitely generated?

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Your argument is correct (though you should mention that you are assuming $v\neq 0$, which implies $\alpha^{-1}(v)\neq 0$ in order to get your inequality). You still need to show that $\alpha^{-1}$ is self-adjoint, though. You can do so by an argument very similar to the one you used.

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  • $\begingroup$ I'm not sure if I can prove it, for $V=\mathbb{C}\:\:or\:\:\mathbb{R}$, my argument is true, but how about when $V$ is infinitely generated inner product space and not equal $\mathbb{C}\:\:or\:\:\mathbb{R}$. And plus in general we don't have $(\alpha^{-1})^* =(\alpha^*)^{-1}$ $\endgroup$
    – Parisina
    Commented Dec 30, 2016 at 6:16
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    $\begingroup$ Just prove directly that $\langle \alpha^{-1}(v),w\rangle=\langle v,\alpha^{-1}(w)\rangle$. The proof is nearly identical to your proof that $\langle \alpha^{-1}(v),v\rangle>0$. $\endgroup$ Commented Dec 30, 2016 at 22:22
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    $\begingroup$ $\langle \alpha^{-1}(v),w\rangle=\langle \alpha^{-1}(v),\alpha\alpha^{-1}(w)\rangle=\langle\alpha\alpha^{-1}(v),\alpha^{-1}(w)\rangle=\langle v,\alpha^{-1}(w)\rangle.$ $\endgroup$
    – Parisina
    Commented Dec 31, 2016 at 0:56
  • $\begingroup$ You know why I got confused,because of this example: Let $V=\mathbb{R}^{(\infty)}$ with inner product $\langle [a_0,a_1,\dots],[b_0,b_1,\dots\rangle=\sum_{i=0}^{\infty}a_ib_i$, Let $\alpha\in End(V)$ be given by $\alpha: [a_0,a_1,dots]\to [0,a_1,a_2,\dots]$. Then $\alpha^*$ exists and is given by $\alpha^*:[a_0,a_1,\dots]\to [a_1,a_2,\dots].$ Here, $\alpha$ is monomorphism but not $\alpha^*.$ $\endgroup$
    – Parisina
    Commented Dec 31, 2016 at 1:04
  • $\begingroup$ But, in this example $\alpha$ isn't an automorphism. Thanks! $\endgroup$
    – Parisina
    Commented Dec 31, 2016 at 1:07

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