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Let $V$ be an finite-dimensional inner product vector space and let $T:V\to V$ be a linear operator such that $T^*=-T$.

Prove that for all $\alpha \in \mathbb{R}$, $I-\alpha T$ is invertible.

My way: I want to show that if $\lambda$ is an eigenvalue of $I-\alpha T$ then $\lambda \ne 0$.

Let $\lambda$ is an eigenvalue with eigenvector $v\ne 0$: $\lambda \langle v,v\rangle =\langle\lambda v,v\rangle=\langle(I-\alpha T)v,v\rangle=\langle v-\alpha Tv,v\rangle=\langle v,v\rangle-\alpha\langle Tv,v\rangle$ $=\langle v,v\rangle+\alpha\langle-Tv,v\rangle=\langle v,v\rangle+\alpha\langle T^{*}v,v\rangle=\langle v,v\rangle+\alpha\langle v,Tv\rangle$

But now $\lambda=\frac{\langle v,v\rangle+\alpha\langle v,Tv\rangle}{\langle v,v\rangle}$ and I don't know if it can be zero

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  • $\begingroup$ Since $T^*=-T$, $\langle Tv,\,v\rangle$ is always in $i\mathbb{R}$. $\endgroup$
    – Aphelli
    Jul 24, 2019 at 12:22

1 Answer 1

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If $\alpha =0$ there is nothing to prove. Otherwise, $I-\alpha T$ not invertible implies that $\beta =\frac 1 {\alpha}$ is an eigen value. Let $Tx=\beta x$ with $x \neq 0$. Then $ \langle Tx, x \rangle =\beta \|x\|^{2}$ and $ \langle Tx, x \rangle =\langle x, T^{*}x \rangle =-\beta \langle x, x \rangle =-\beta \|x\|^{2}$. We have arrived at a contradiction.

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