For a bounded linear operator $T$ and $\alpha\in\mathbb{C}$, show the spectrum $\sigma(\alpha T)=\alpha\sigma(T)$

Question

$\sigma(\alpha T)=\{\lambda\in\mathbb{C}\mid \alpha T-\lambda I\text{ not invertible}\}$

$\alpha\sigma(T)=\{\alpha\gamma\mid\gamma\in\sigma(T)\}=\{\alpha\gamma\mid T-\gamma I\text{ not invertible}\}$

I want to show $\sigma(\alpha T)=\alpha\sigma(T)$

Of course the natural way would be showing each set is contained in the other. So I want to show that if I have $\lambda$ which makes $\alpha T-\lambda I$ non-invertible, then this $\lambda =\alpha\gamma$ where $\gamma$ makes $T-\gamma I$ non-invertible.

But my main issue is I don't really have a good grasp of the properties of inverse operators.

Eg I'd like to say something like $\displaystyle{\alpha T-\lambda I =\alpha T-\frac{\lambda}{\alpha}\alpha I \text{ not invertible } \implies T-\frac{\lambda}{\alpha}\alpha I\text{ not invertible } \implies \gamma =\frac{\lambda}{\alpha}}$,

but I have no idea if that follows.

Answer 1

Let $\alpha\ne0$. \begin{align*}\lambda\notin\sigma(\alpha T)&\iff \alpha T-\lambda\textrm{ is invertible}\\ &\iff \alpha(T-\lambda/\alpha)\textrm{ is invertible}\\ &\iff (T-\lambda/\alpha)\textrm{ is invertible}\\ &\iff \lambda/\alpha\notin\sigma(T)\\ &\iff \lambda\notin\alpha \sigma(T) \end{align*}

So $\sigma(\alpha T)=\alpha\sigma(T)$. Can you complete the proof for $\alpha=0$?