# How do I construct an operator with a very specific spectrum?

I have been studying functional analysis lately, specifically spectrum of operators. I know how to find the spectra of an operator, but what if I have the spectra and I want to find an operator with such spectra?

Let's say I have $$\delta(T) = \{3i\}$$ as the spectra of some linear and bounded operator T. I guess that T could be something like this:

$$T: l^2 \to l^2$$, such that $$\{x\}_j \to 3i\{x\}_j$$ could be a trivial candidate because $$Tx = \lambda x$$. However what if I have $$\delta(T) = \{3i,5i,7i\}$$ for example?

• You can take an orthonormal basis $e_1,e_2,...$ for $l^2$. Multiply $e_1$ by $3i$, $e_2$ by $5i$ and all the rest by $7i$. In other words $(x_1,x_2,x_3,x_4,...)\mapsto (3i x_1,5ix_2,7ix_3, 7ix_4,...)$. – Yanko Feb 25 at 13:56
• @Yanko Can I also just take $(3ix_1, 5ix_2, 7ix_3, x_4,x_5,x_6,...)$? – qcc101 Feb 25 at 14:07
• No, because then $1$ is also in the spectrum. – Yanko Feb 25 at 14:07

It is simple to do so because the multiplication operator $$T: L^2(\mathbb{R}) \to L^2(\mathbb{R})$$, $$f \mapsto g f$$, with some bounded function $$g: \mathbb{R} \to \mathbb{C}$$, has as spectrum the closure of the range of $$g$$.
So for your first example, you take the constant function $$g: g(x) = 3i$$ and get the spectrum $$\{ 3i\}$$. Similarly you can choose any function with $$g(\mathbb{R}) = \{3i,5i,7i\}$$ for your second example (take something peacewise constant for example) or similarly in any example you can think of.
• (+1) I think $l^2$ in this context means $l^2(\mathbb{N})$ or $l^2(\mathbb{Z})$. However the idea is the same. – Yanko Feb 25 at 13:58
• Oh, I was actually thinking about $l^2$ as sequences, is that the same? This one: en.wikipedia.org/wiki/Sequence_space – qcc101 Feb 25 at 14:01
• @qcc101 yes the idea is the same. Just consider $g$ as a function from $\mathbb{N}$ to $\mathbb{C}$ instead. – Yanko Feb 25 at 14:06
• On $L^2(\mathbb{R})$, the spectrum of the multiplication operator is not the closure of the range of $g$, but the essential range (you can change $g$ on a set of measure zero without changing the operator). – MaoWao Feb 25 at 14:06