2B (Vectors and Matrices)

Taking determinants of

M^{T} J M = J

gives

det (M)^{2} det J = det J .

(1)

Since

det J = - 1 \neq = 0

, one has

det (M)^{2} = 1

, and hence

det M = 1 or det M = - 1 .

(2)

Both values occur, for example for

M = I

and

M = diag (1, - 1)

ii

 0  0

Solution

 0  0

M_{1}^{T} J M_{1} = J

and

M_{2}^{T} J M_{2} = J

, then matrix multiplication and the transpose rule give

(M_{1} M_{2})^{T} J (M_{1} M_{2}) = M_{2}^{T} (M_{1}^{T} J M_{1}) M_{2} = M_{2}^{T} J M_{2} = J .

(1)

Thus the product also satisfies the equation.

The equation and part (i) imply that

M_{1}

is invertible. Multiplying

M_{1}^{T} J M_{1} = J

on the left by

(M_{1}^{- 1})^{T}

and on the right by

M_{1}^{- 1}

gives

(M_{1}^{- 1})^{T} J M_{1}^{- 1} = J .

(2)

Hence the set of solutions is closed under products and matrix inverses.

b

 0  0

Solution

 0  0

Write

M = (p r q s) .

(1)

The first column has unit length for the indefinite quadratic form:

p^{2} - r^{2} = 1.

(2)

Because

p = M_{11} > 0

, there is a unique

u \in R

such that

p = cosh u, r = sinh u .

(3)

The two columns are orthogonal for the same form and the second has squared length

- 1

pq - rs = 0, q^{2} - s^{2} = - 1.

(4)

The vectors with these properties are

(q, s) = \pm (sinh u, cosh u) .

(5)

Using the hyperbolic cosine and hyperbolic sine, the two possible families are therefore

(cosh u sinh u sinh u cosh u), (cosh u sinh u - sinh u - cosh u), u \in R .

(6)

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 2B (Vectors and Matrices)

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