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I'm working on an exercise from Steven Roman's Advanced Linear Algebra. He asks

Let $\dim(V)<\infty$. If $\tau,\sigma\in\mathscr{L}(V)$, prove that $\sigma\tau=\mathrm{id}$ implies $\tau$ and $\sigma$ are invertible, and that $\sigma=p(\tau)$ for some polynomial $p(x)\in F[x]$, $F$ being the underlying field.

From $\sigma\tau=\mathrm{id}$, $\tau$ must be injective, hence surjective, and thus invertible. Thus $\sigma=\tau^{-1}$, so both $\sigma$ and $\tau$ are invertible. But it is not clear to me how $\sigma$ can be expressed as a polynomial in $\tau$. Can anyone please explain?

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As $\dim(V) < \infty$, $\dim \mathscr L(V) < \infty$. So there are $a_i \in F$, not all $a_i = 0$, such that $\sum_{i=0}^n a_i\tau^i = 0$. As $\tau$ is invertible, we may suppose that $a_0 \ne 0$ (otherwise let $k = \min\{i\mid a_i \ne 0\}$ and multiply by $\tau^{-k}$). So $$ \mathrm{id} = -\frac 1{a_0}\sum_{i=1}^n a_i \tau^i = \tau \circ \sum_{i=0}^{n-1} \frac{-a_{i+1}}{a_0}\tau^i $$ So $$ \tau^{-1} = \sum_{i=0}^{n-1} \frac{-a_{i+1}}{a_0}\tau^i $$ now let $p(x) = \sum_{i=0}^{n-1} \frac{-a_{i+1}}{a_0}x^i$ giving $p(\tau) = \tau^{-1} = \sigma$.

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  • $\begingroup$ Perfectly clear now, thanks. $\endgroup$ Commented Nov 13, 2012 at 8:25

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