Is $0$ an eigenvalue of a compact normal operator? Is $0$ an eigenvalue for a compact normal operator?
Many texts mention that compact normal operators have a complete orthonormal basis of eigenvectors. If they do, then what about the kernel of the operator? The elements in the kernel, may not be eigenvectors.
Where is the mistake in my understanding?
 A: A compact normal operator need not have zero as an eigenvalue. But nothing rules that possibility out.
For easy examples, we have the operator $A$ represented by the following matrix in the standard basis on $\mathbb{R}^2$:
$$
[A] = \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}
$$
Then $A:\mathbb{R}^2\to\mathbb{R}^2$ is a Hermitian operator, hence normal. It is also continuous and compact, because all linear operators on finite-dimensional normed vector spaces are continuous and compact. Finally, it has $1$ and $0$ as eigenvalues, and an obvious orthonormal basis of eigenvectors.
On the other hand, the identity $I:\mathbb{R}^2\to\mathbb{R}^2$ is also normal, continuous, and compact, but does not have $0$ as an eigenvalue.
A: The elements of the kernel are precisely the eigenvectors for zero.
So, when $T $ is normal and compact you can form an orthonormal basis by joining orthonormal bases for each eigenspace (including the kernel, if nonzero).
A: If $X$ is a normed space and $\dim X < \infty$, then each linear operator $K:X \to X$ is compact. Thus $0$ may or may not be an eigenvalue.
