 Solution

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

Put

n = dim V > 0

and

r = rank T

. The direct sum in part (i) gives

χ_{T} (x) = x^{n - r} (x - 1)^{r} .

The eigenvalues are the members of

{0, 1}

actually present: zero occurs exactly when

r < n

, and one exactly when

r > 0

. Since

T

m_{T} (x) = ⎩ ⎨ ⎧ x, x - 1, x (x - 1), T = 0, T = I, T \neq = 0, I .

In particular, a nonzero vector space does not force both eigenvalues to occur. The zero and identity maps are important special cases.

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