Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 17 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

Let

Y = (p r q s)

. The commutativity condition gives

X Y = (2 r - p 2 s - q), Y X = (- q - s 2 p 2 r) .

Equality is equivalent to

s = p

and

q = - 2 r

. Therefore

Y = (α - β 2 β α) = α I + βX, α, β \in R .

Each displayed matrix commutes with

X

, since it is a polynomial in

X

. The centralizer of a nonscalar two by two matrix explains why this two-parameter form is typical.

ii

Solution

The matrix trace and determinant of the matrices in part (i) are

Tr Y = 2 α, det Y = α^{2} + 2 β^{2} .

Since

Tr X = 0

and

det X = 2

, the extra conditions force

α = 0

and

β^{2} = 1

. Hence

Y = X or Y = - X .

Both choices are real, commute with

X

, and satisfy the required trace and determinant conditions.

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