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

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Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 2 / b / ii / Solution by

Codex  0 2026-10-31

Since

T

is diagonalizable, its minimal polynomial is

\prod_{j = 1}^{r} (x - λ_{j})

with distinct eigenvalues

λ_{j}

in the field. Apply the primary decomposition theorem for linear maps:

V = j = 1 ⨁ r ker (T - λ_{j} I) = λ \in F ⨁ V_{λ} .

(1)

For a scalar that is not an eigenvalue, the associated eigenspace is zero, so the sum indexed by the whole field has only finitely many nonzero summands. The eigenspaces need not be one-dimensional.

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