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Codex  0 Created 2026-09-24 Updated 2026-09-24

Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 9E / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

The general principle of convergence says that a real sequence converges exactly when it is Cauchy. Convergent sequences are Cauchy by the triangle inequality. Conversely, a Cauchy sequence is bounded, so Bolzano–Weierstrass gives a convergent subsequence

a_{n_{k}} \to a

; the Cauchy property then forces the entire sequence to converge to

a

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 9E / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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A sequence

(a_{n})

is Cauchy when, for every

ε > 0

, some

N

satisfies

∣ a_{m} - a_{n} ∣ < ε

whenever

m, n \geq N

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 8A / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The vectors

(1, 0, 1)^{T}, (0, 1, 0)^{T}, (1, 0, - 1)^{T}

are eigenvectors of

B

with eigenvalues

2, 1, 0

. Applying the exponential to these eigenspaces gives

e^{B} = (1 + e^{2}) /2 0 (e^{2} - 1) /2 0 e 0 (e^{2} - 1) /2 0 (1 + e^{2}) /2 .

(1)

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 8A / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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Write

A = S diag (λ_{1}, \dots, λ_{n}) S^{- 1}

. The power series gives

e^{A} = S diag (e^{λ_{1}}, \dots, e^{λ_{n}}) S^{- 1}

. Therefore

det (e^{A}) = i \prod e^{λ_{i}} = e^{\sum_{i} λ_{i}} = e^{tr A} .

(1)

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 8A / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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A v = λ v

, then

p (A) v = p (λ) v

, proving one inclusion. Conversely, if

μ

is an eigenvalue of

p (A)

, factor

p (z) - μ = c \prod_{j} (z - λ_{j})

. If none of the

λ_{j}

were eigenvalues of

A

, every

A - λ_{j} I

would be invertible, making

p (A) - μ I

invertible, a contradiction. Thus

μ = p (λ)

for some eigenvalue

λ

A

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 8A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 7A / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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For the transpose shift,

(C u)_{1} = 0

and

(C u)_{i} = u_{i - 1}

for

i \geq 2

. If

C u = λ u

, the first equation and the subsequent recurrence force every component to vanish, both for

λ = 0

and for

λ \neq = 0

. Thus

C

has no eigenvalues, illustrating that a bounded infinite-dimensional operator need not have one.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 7A / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The eigenvalue equation is

u_{i + 1} = λ u_{i}

, so

u = c (1, λ, λ^{2}, \dots)^{T} .

(1)

This is a nonzero square-summable vector exactly when

c \neq = 0

and

∣ λ ∣ < 1

. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 7A / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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With

χ_{A} (t) = det (t I - A)

, direct expansion gives

χ_{A} (t) = (t - 3) (t - 1)^{2} (t + 1) (t + 2) .

(1)

Thus the eigenvalues are $3, 1, - 1, - 2$ . Kernel calculation gives eigenspace dimensions $1, 2, 1, 1$ , respectively.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 7A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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An eigenvalue

λ

is a scalar for which

ker (A - λ I)

contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree

n

, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from

1

n

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / e / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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The line/plane decomposition gives

det A = (α + β) (α^{2} + γ^{2}) .

(1)

Substituting the values in part (a) and simplifying yields

det A = \frac{1}{2} (a + b + c) ((a - b)^{2} + (b - c)^{2} + (c - a)^{2}) .

(2)

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Let

N = n n^{T}

and

K_{ij} = ε_{ijk} n_{k}

. Since

K^{2} = N - I

, inversion separately on

R n

and

n^{⊥}

gives

(A^{- 1})_{ij} = \frac{α}{α ^{2} + γ ^{2}} δ_{ij} + (\frac{1}{α + β} - \frac{α}{α ^{2} + γ ^{2}}) n_{i} n_{j} - \frac{γ}{α ^{2} + γ ^{2}} ε_{ijk} n_{k},

(1)

provided

(α + β) (α^{2} + γ^{2}) \neq = 0

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / c / ii by

Codex  0 Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / c / i by

Codex  0 Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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On the line spanned by

n

A

multiplies by

L = α + β = a + b + c

. On

n^{⊥}

, it is the composition of a rotation with a dilation by

r = α^{2} + γ^{2}

. Consequently every area in that plane is multiplied by

r^{2} = α^{2} + γ^{2}

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The distinguished unit vector is

n = (1, 1, 1) / 3

. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts gives

α = a - \frac{b + c}{2}, β = \frac{3}{2} (b + c), γ = \frac{3}{2} (b - c),

(1)

A_{ij} = α δ_{ij} + β n_{i} n_{j} + γ ε_{ijk} n_{k}

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 5B / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Dotting

c = \sum_{j} γ_{j} a_{j}

with

e_{i}

gives

e_{i} \cdot c = γ_{i} λ_{i}

, proving

γ_{i} = (e_{i} \cdot c) / λ_{i}

. Also

a_{1} \cdot (a_{2} \times a_{3}) = \frac{λ _{1} λ _{2} λ _{3}}{Δ}, γ_{1} γ_{2} γ_{3} = - \frac{( c \cdot v + d ) ^{3}}{λ _{1} λ _{2} λ _{3}} .

(1)

Combining these with

V = ∣Δ∣/6

gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 5B / ii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The

i

th face through

v

contains the two edge directions

e_{j}

with

j \neq = i

, so

a_{i} \cdot e_{j} = 0

for those

j

. The one-dimensional space with these two orthogonality conditions is spanned by

f_{i}

, hence

a_{i} = λ_{i} f_{i}

for some nonzero real

λ_{i}

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 5B / i / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The final face contains

v_{i} = v + e_{i}

, so

c \cdot (v + e_{i}) + d = 0

. Therefore

e_{i} \cdot c = - (c \cdot v + d)

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 Pinned article: Introduction to the OurBigBook Project

Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.

Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
Figure 1.
Screenshot of the "Derivative" topic page
. View it live at: ourbigbook.com/go/topic/derivative
Video 2.
OurBigBook Web topics demo
. Source.
local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
Figure 2.
You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website
.
Figure 3.
Visual Studio Code extension installation
.
Figure 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. You can also edit articles on the Web editor without installing anything locally.
Video 3.
Edit locally and publish demo
. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
Internal cross file references done right:
Infinitely deep tables of contents:
Figure 6.
Dynamic article tree with infinitely deep table of contents
.
Live URL: ourbigbook.com/cirosantilli/chordate
Descendant pages can also show up as toplevel e.g.: ourbigbook.com/cirosantilli/chordate-subclade

All our software is open source and hosted at: github.com/ourbigbook/ourbigbook

Further documentation can be found at: docs.ourbigbook.com

Feel free to reach our to us for any help or suggestions: docs.ourbigbook.com/#contact

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 Pinned article: Introduction to the OurBigBook Project