Is $Ax = (\alpha_1 x_1, \alpha_2 x_2, \alpha_3 x_3, \dots, \alpha_k x_k, 0, 0, \dots)$ a compact operator?

Question

Is the operator $A$ defined by

$$Ax = (\alpha_1 x_1, \alpha_2 x_2, \alpha_3 x_3, \dots, \alpha_k x_k, 0, 0, \dots)$$

a compact operator?

It only has finitely non-zero dimensions, so does this mean it can be considered to be of finite dimension and hence compact?

Answer 1

This depends on the domain and range of the the operator $A$.

For a linear operator $A : X \to Y$, where $X$ and $Y$ are normed spaces, one can show that if $A$ is bounded and of finite rank, i.e., $\dim A(X) < \infty$, then the operator $A$ is compact.

So, if you consider normed spaces, then your reasoning is fine.