 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/2/c/solution

Let

λ

v_{1}, \dots, v_{n}

of eigenvectors, and every corresponding eigenvalue is

λ

. For any

v = \sum_{i} a_{i} v_{i}

T (v) = \sum_{i} a_{i} T (v_{i}) = \sum_{i} a_{i} λ v_{i} = λ v .

Thus

T = λ I

J = (2012) .

(t - 2)^{2}

, but

(J - 2 I) (a, b)^{T} = (b, 0)^{T}

. The eigenspace for

2

is therefore

span {(1, 0)^{T}}

, of dimension one. There is no basis of eigenvectors in dimension two, so

J

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