Let's denote the vector space \( V \) of polynomials of degree at most 3. A general polynomial in \( V \) can be expressed in the form:
\[ f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0 \]
where \( a_3, a_2, a_1, a_0 \) are constants from the field (for example, real or complex numbers).
The differentiation operator \( T: V \to V \) acts on a polynomial \( f \) by differentiating it:
\[ T(f) = f' = 3a_3 x^2 + 2a_2 x + a_1 \]
The kernel of the operator \( T \), denoted \( \ker(T) \), is the set of polynomials \( f \in V \) such that \( T(f) = f' = 0 \).
The derivative of a polynomial \( f \) is equal to zero if and only if \( f \) is a constant polynomial. This is because the only polynomial whose derivative is zero is a constant polynomial (i.e., a polynomial of degree 0).
Polynomials of degree at most 3 that are constant can be expressed in the form:
\[ f(x) = a_0 \]
where \( a_0 \) is a constant (with \( a_1 = 0, a_2 = 0, a_3 = 0 \)).
The space of constant polynomials (polynomials of degree 0) is a one-dimensional vector space. Thus, we can conclude that:
\[ \dim \ker(T) = 1 \]
The dimension of the kernel of the differentiation operator \( T \) is therefore \( \boxed{1} \).
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