Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 7B (Vectors and Matrices)

Table of contents
- a 7B
  - Solution a
- b 7B
  - Solution b
- c 7B
  - Solution c

a

Solution

For

x \neq = 0

,

x^{†} G x = ∣ A x ∣^{2} > 0,

so the Hermitian matrix

G

has positive real eigenvalues. If

G e_{i} = λ_{i} e_{i}

, then

H (A e_{i}) = A A^{†} A e_{i} = A (G e_{i}) = λ_{i} A e_{i} .

Moreover

∣ A e_{i} ∣^{2} = λ_{i} ∣ e_{i} ∣^{2}

, so

∣ f_{i} ∣/∣ e_{i} ∣ = λ_{i}

.

b

Solution

Choose an orthonormal eigenbasis

e_{i}

of

G

and set

u_{i} = e_{i}

,

v_{i} = A e_{i} / λ_{i}

. Part (a) shows that the

v_{i}

form an orthonormal eigenbasis of

H

. If

U

and

V

have these vectors as columns, then they are unitary and

V^{†} A U = diag (λ_{1}, \dots, λ_{n}) .

This is the singular value decomposition.

c

Solution

Here

A^{†} A = \frac{1}{2} (5 - 3 - 3 5),

with eigenvalues

4, 1

and corresponding normalized vectors

(1, - 1)^{T} / 2

and

(1, 1)^{T} / 2

. A compatible choice is

U = \frac{1}{2} (1 - 1 11), V = (10 0 - 1), D = (2001) .

Direct multiplication gives

V^{†} A U = D

.

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