Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 6 / a

Table of contents
- Solution a
- i a
  - Solution i
- ii a
  - Solution ii

Solution

Under an orthogonal Cartesian change of axes

Q

, the vector components transform as

v_{i}^{'} = Q_{ia} v_{a}

, while

v_{k}^{'} v_{k}^{'} = v_{a} v_{a}

and

δ_{ij} = Q_{ia} Q_{ja}

. Substitution gives

P_{ij}^{'} = δ_{ij} - \frac{Q _{ia} v _{a} Q _{jb} v _{b}}{v _{c} v _{c}} = Q_{ia} Q_{jb} (δ_{ab} - \frac{v _{a} v _{b}}{v _{c} v _{c}}) = Q_{ia} Q_{jb} P_{ab} .

This is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form

P = I - v v^{T} / (v^{T} v)

is the orthogonal projection matrix onto the orthogonal complement of

v

.

i

Solution

Contract with

v_{j}

and use its squared Euclidean norm:

P_{ij} v_{j} = v_{i} - v_{i} \frac{v _{j} v _{j}}{v _{k} v _{k}} = 0.

Thus

v

lies in the kernel of this orthogonal projection.

ii

Solution

The orthogonality condition is

v_{j} u_{j} = 0

, so

P_{ij} u_{j} = u_{i} - v_{i} \frac{v _{j} u _{j}}{v _{k} v _{k}} = u_{i} .

Every vector in the orthogonal complement is fixed by the orthogonal projection.

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