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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$.

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    $\begingroup$ yes, that's correct $\endgroup$
    – Exodd
    Commented May 5, 2022 at 20:57
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    $\begingroup$ 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 $-1$. $\endgroup$
    – Nate
    Commented May 5, 2022 at 21:01
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    $\begingroup$ 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! $\endgroup$ Commented May 5, 2022 at 21:09
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    $\begingroup$ 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 $\forall v~((T-I)v=0)$ $\endgroup$ Commented May 5, 2022 at 22:57
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    $\begingroup$ @DarbyBond I prefer to argue that way: as $(T+I)(T-I)v=0,$ then $(T-I)v=0.$ Since $v$ is arbitrary $T-I=0.$ $\endgroup$ Commented May 6, 2022 at 2:26

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