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

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

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

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.