Proof of $\sigma(zA) = z\sigma(A)$

Question

Let $\mathfrak{U}$ be a $C^\star$-algebra with unity $I$ and let $\sigma(A)$ be the spectrum of an operator $A \in \mathfrak{U}$.

$\sigma(A)$ is the set of all complex numbers $\lambda$ such that $A-I\lambda$ has not (two-sided) inverse.

I want to prove that, for a $z \in \mathbb{C}$, we have that

$$\sigma(zA) = z\sigma(A)$$

Where $z\sigma(A):= \{z\lambda \in \mathbb{C}: \lambda \in \sigma(A)\}$.

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My attempts: Let $x \in z\sigma(A):= \{z\lambda \in \mathbb{C}: \lambda \in \sigma(A)\}$ this implies that $x = z\lambda_x$ for some $\lambda_x \in \sigma(A)$, wich is the same as saying that $A - \lambda_x I$ has not an inverse. Suppose $z \neq 0$ then this implies that:

$A - (x/z)I$ doesn't have an inverse iff

$ z^{-1}(zA-xI)$ doesn't have an inverse iff

$zA-xI$ doesn't have an inverse.

So $x \in \sigma(zA)$. Then $z\sigma(A)\subset\sigma(zA)$.

Let now $x \in \sigma(zA)$ then we have the situation before stated in such a way that if $(zA-xI)$ does not have an inverse then $x/z := \lambda \implies x = z\lambda$ where $\lambda \in \sigma(A)$ and we get our solution.

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Questions: what if $z=0$? This would let $x = 0$ and something like $\sigma(zA=0)$? There is any problem in my proof? Can someone provide a rigorous (hints of a ) demonstration? This is important in proving a theorem in quantum mechanics. The second part of the proof (if correct) seems dispensable to me because the first part seems like iff cases only, is this true?

Answer 1

The answer by Andre' is the same as the Rafael's proof and is only valid for the case $z \ne 0$, because the assertion $$(A - \lambda I) \ \ \text{ is invertible} \ \ \iff z(A - \lambda I) \ \ \text{ is invertible}$$ is valid only if $z \ne 0$.

Let's pass to the case $z = 0$, which was the focus of the question. In this case $z A = 0$, the zero element of the $C^*$ algebra, and the spectrum of the zero operator is $\{0\}$, because $0 - \lambda I = - \lambda I$ is invertible for $\lambda \ne 0$. On the other hand, and this is the only place that some analysis is used, it is a basic theorem of functional analysis that for any $A$, $\sigma(A)$ is a non-empty compact subset of $\mathbb C$. Hence $0 \sigma(A) = \{0\}$.

Answer 2

$$\lambda\in \sigma(A) \Leftrightarrow A-\lambda I \textrm{ has no inverse}\Leftrightarrow z(A-\lambda I) \textrm{ has no inverse}\Leftrightarrow zA-z\lambda I \textrm{ has no inverse};$$ $$\Leftrightarrow z\lambda\in \sigma(zA)$$ (With $z\neq 0$, otherwise $z(A-\lambda I)$ is alwais non-invertible - regardless of A properties)

Then, $\alpha \in\sigma(zA)$ if, and only if, there is a $\lambda\in \sigma (A)$ s.t. $\alpha=z\lambda$. I.e., iff $\alpha \in z\sigma(A)$. So every element of $\sigma(zA)$ is an element of $z\sigma(A)$ and vice-versa.