Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 6C / ii

Solution

Let

e_{1}, \dots, e_{n}

be the standard basis and define

a_{j} = f (e_{j})

f (x) = f (\sum_{j = 1}^{n} x_{j} e_{j}) = \sum_{j = 1}^{n} x_{j} f (e_{j}) = a \cdot x .

Thus every linear functional has this form. Its representing vector is unique because evaluating

a \cdot x = b \cdot x

at each

e_{j}

gives

a_{j} = b_{j}

For the prescribed kernel of a linear map, the third listed vector is the first minus the second, so it adds no independent constraint. The first two require

a_{1} + a_{2} + a_{3} - a_{4} = 0, 2 a_{1} - a_{2} - 2 a_{4} = 0.

Taking

a_{1} = s

and

a_{4} = t

gives

a_{2} = 2 s - 2 t

a_{3} = - 3 s + 3 t

. Consequently all possible representing vectors are

a = s (1, 2, - 3, 0)^{T} + t (0, - 2, 3, 1)^{T}, s, t \in R .

Conversely, every such vector is perpendicular to each of the three prescribed vectors, so its linear functional vanishes on them. This also includes the zero linear functional.

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