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My attempt:

Since $ -1$ is not an eigenvalue of $T\implies T+I$ is invertible, that is $(T+I)^{-1}$ exists. Now $(T+I)(T-I) = T^2 - I \implies (T+I)(T-I) = 0$.

After applying $(T+I)^{-1 }$ on left and right hand side of the previous equation we get: $ (T+I)^{-1} (T+I) (T-I) = (T+I)^{-1}(0) \implies T-I = 0 \implies T = I$.

Yuval Peres
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Darby Bond
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    yes, that's correct – Exodd May 05 '22 at 20:57
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    Your proof looks fine to me. Here is another way to see it:

    Suppose $T(\vec{v}) \ne \vec{v}$ for some vector $\vec{v}$, then $T(\vec{v}) - \vec{v} \ne 0$ is a eigenvector with eigenvalue $-1$.

     – Nate May 05 '22 at 21:01
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    Another way: since $T^2-I=0$, the minimal polynomial of $T$ divides $x^2-1=(x+1)(x-1)$. But the roots of the minimal polynomial are the eigenvalues of $T$; since $-1$ is not an eigenvalue, the minimal polynomial must actually divide $x-1$, so that $T-I=0$. This illustrates a general method—but that being said, in this example I like your solution better! – Greg Martin May 05 '22 at 21:09
  • I would argue as follows. If $-1$ is not an eigenvalue then $I+T$ is injective. Hence $(T+I)(T-I)=0$ implies $T-I=0.$ – Ryszard Szwarc May 05 '22 at 21:29
  • @RyszardSzwarc could you expand on your argument? I get that $T-I$ is not injective but how do we conclude $T-I=0$? – Darby Bond May 05 '22 at 22:09
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    If $(T-I)v \neq 0$ for some $v$, then $T+I$ injective implies $(T+I)((T-I)v) \neq 0$. By assumption, that can't happen, so $\forall v~((T-I)v=0)$ – Robert Shore May 05 '22 at 22:57
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    @DarbyBond I prefer to argue that way: as $(T+I)(T-I)v=0,$ then $(T-I)v=0.$ Since $v$ is arbitrary $T-I=0.$ – Ryszard Szwarc May 06 '22 at 02:26
  • Wow very concise – Darby Bond May 06 '22 at 04:11

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