 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/2/b/i/solution

m_{T} (x) = \prod_{j = 1}^{r} p_{j} (x)^{e_{j}}

p_{j}

, then

V = j = 1 ⨁ r ker (p_{j} (T)^{e_{j}}) .

Each summand is a

T

-invariant subspace. More generally the same direct sum conclusion holds for any factorization of an annihilating polynomial into pairwise coprime polynomials, using the corresponding kernels. The underlying mechanism is the Bézout's identity for polynomials, which separates the mutually coprime factors.

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