 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/calculus-probability-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/3/b/ii/solution

T (x, y) = (y, 0), [T] = (0010) .

(x, 0)

, and its image of a linear map is the same line:

ker T = im T = span {(1, 0)} .

Their set intersection is nonzero and their sum is only one-dimensional. In particular,

(0, 1)

cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. Thus

R^{2} \neq = ker T \oplus im T .

The matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

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