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Recall the following definitions and properties: 1. A linear operator T on a vector space V is a function that maps elements of V to elements of V and preserves vector addition and scalar multiplication. 2. A subspace W of V is called T-invariant if T(w) is in W for all w in W. 3. The minimal polynomial of a linear operator T is a polynomial p(t) with coefficients in the underlying field (usually real or complex numbers) such that p(T) = 0, where 0 represents the zero linear transformation. The minimal polynomial is the monic polynomial of minimal degree that satisfies this condition. 4. The direct sum of two subspaces W1 and W2, denoted as W1 ⊕ W2, is the set of all sums of element from W1 and W2, and in this case, it generates the entire vector space V. This implies that W1 and W2 have no non-zero intersections, i.e., W1 ∩ W2 = {0}. Now, we aim to either prove that the minimal polynomial f(t) of T is equal to the product of the minimal polynomials p1(t) and p2(t) of T_W1 and T_W2, respectively, or find a counterexample to disprove this claim.

Suppose that f(t) = p1(t)p2(t). Then, f(T) = p1(T)p2(T), and for any vector v in V, we have: f(T)(v) = p1(T)p2(T)(v) Now, since every vector v in V can be written as v = w1 + w2 with w1 in W1 and w2 in W2 due to the direct sum construction, we have: f(T)(v) = p1(T)p2(T)(w1 + w2) By considering the T-invariant property of W1 and W2, we can analyze if f(T)(v) = 0 for all v in V. If that is the case, then by the definition of minimal polynomials, f(t) must be the minimal polynomial of T.

Consider V = R^2, the vector space of all column vectors with 2 real entries, and T: V → V defined as T(x, y) = (x, 0) This linear operator T has two T-invariant subspaces, namely W1 = {(x, 0) | x ∈ R} and W2 = {(0, y) | y ∈ R}. Now, the restrictions of T to W1 and W2 are the respective operators T_W1: W1 → W1 and T_W2: W2 → W2 defined as follows: T_W1(x, 0) = (x, 0) T_W2(0, y) = (0, 0) The minimal polynomials of T_W1 and T_W2 are p1(t) = t - 1 and p2(t) = t, respectively. Thus, the product p1(t)p2(t) = t^2 - t. Now, consider the minimal polynomial of T. It must satisfy two conditions: f(T) = 0, and f(t) is of minimal degree. We can see that the polynomial f(t) = t(t - 1) = t^2 - t satisfies both conditions since f(T) maps any vector in V to the zero vector and f(t) is of minimal degree 2. Therefore, we can see that f(t) = p1(t)p2(t) in this case, and the given claim is true. In conclusion, the minimal polynomial f(t) of T is indeed equal to the product of the minimal polynomials p1(t) and p2(t) of T_W1 and T_W2 when V is the direct sum of T-invariant subspaces W1 and W2.