Does an eigenvalue that does NOT have multiplicity usually have a one-dimensional corresponding eigenspace? I'm trying to understand this statement in my book:

In general, the multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.

So then, if an eigenvalue does NOT occur as a multiple root, is that kind of like saying it has a multiplicity of one and therefore the dimension of its eigenspace is less than or equal to one?
I guess I'm just struggling to understand the concept of multiplicity and why it matters, and what having multiple roots of the characteristic polynomial even means or how it affects anything.

The eigenspace of a particular eigenvalue is guaranteed to have dimension at least $1$. This is because in finding the eigenvalues of a matrix $A$, we require $\det(A-\lambda I)=0$, which guarantees that the system $(A-\lambda I)\mathbf{x}=\mathbf{0}$ will have a nontrivial solution. So there will be at least one eigenvector which satisfies $A\mathbf{x}=\lambda\mathbf{x}$ for each eigenvalue $\lambda$.
The issue when an eigenvalue $\lambda$ has multiplicity greater than $1$ is that we cannot guarantee that the eigenspace will have the same dimension as the multiplicity. We can only guarantee that the eigenspace will have dimension at least $1$.
In particular, it tells us about diagonalizability of a matrix. If an $n\times n$ matrix $A$ has $n$ distinct eigenvalues (i.e. each eigenvalue has multiplicity $1$), then it is diagonalizable. It may still be diagonalizable if it doesn't have $n$ distinct eigenvalues, but there is no guarantee.
